defs-CBB+CBA-sol.tex

%%!TEX root = the-defs-CBB+CBA-sol.tex
\parbf{\ref{exr-crofton}.}
Suppose $\alpha$ is a closed spherical curve. 
By Crofton's formula, the length of  $\alpha$  is $\pi\cdot n_\alpha$, where $n_\alpha$ denotes the average number of crossings of $\alpha$ with equators.

Since $\alpha$ is closed, almost all equators cross it at an even number of points (we assume that $\infty$ is an even number).
If $\length \alpha<2\cdot\pi$, then $n_\alpha<2$.
Therefore there is an equator that does not cross $\alpha$ --- hence the result.


\parbf{\ref{ex:complete=>complete}}; \ref{SHORT.ex:complete=>complete:complete}.
Note that any Cauchy sequence $x_n$ in $(\spc{X},\yetdist{}{}{})$ is also Cauchy in $\spc{X}$.
Since $\spc{X}$ is complete, $x_n$ converges; denote its limit by $x_\infty$.

Passing to a subsequence, we may assume that $\yetdist{x_{n-1}}{x_n}{}<\tfrac1{2^n}$.
It follows that there is a 1-Lipschitz curve $\alpha\:[0,1]\to (\spc{X},\yetdist{}{}{})$ such that $x_n=\alpha(\tfrac1{2^n})$ and $x_\infty=\alpha(0)$.
In particular, $\yetdist{x_n}{x_\infty}{}\to0$ and $n\to\infty$.

\parit{\ref{SHORT.ex:complete=>complete:compact}.}
Fix two points $x,y\in\spc{X}$ such that $\ell=\yetdist{x}{y}{}<\infty$.
Let $\alpha_n$ be a sequence of paths from $x$ to $y$ such that $\length(\alpha_n)\to\ell$ as $n\to \infty$.
Without loss of generality, we may assume that each $\alpha_n$ is $(\ell+1)$-Lipschitz.

Since $\spc{X}$ is compact, there is a partial limit $\alpha_\infty$ of $\alpha_n$ as $n\to \infty$.
By semicontinuity of length, $\length\alpha_\infty\le\ell$;
that is; $\alpha$ is a shortest path in~$\spc{X}$.

\parit{Source:} Part \ref{SHORT.ex:complete=>complete:complete} appears as a Corollary in \cite{hu-kirk}; see also \cite[Lemma 2.3]{petrunin-stadler}.

\parbf{\ref{ex:no-geod}.}
The following example was suggested by Fedor Nazarov~\cite{nazarov}.

\medskip

Consider the unit ball $(B,\rho_0)$
in the space $c_0$ of all sequences converging to zero equipped with the sup-norm.

Consider another metric $\rho_1$ which is different from $\rho_0$ by the conformal factor
\[\phi(\bm{x})=2+\tfrac{1}2\cdot x_1+\tfrac{1}4\cdot x_2+\tfrac{1}8\cdot x_3+\dots,\]
where $\bm{x}=(x_1,x_2\,\dots)\in B$.
That is, if $\bm{x}(t)$, $t\in[0,\ell]$, is a curve parametrized by $\rho_0$-length, then its $\rho_1$-length is 
\[\length_{\rho_1}\bm{x}=\int\limits_0^\ell\phi\circ\bm{x}.\]
Note that the metric $\rho_1$ is bi-Lipschitz equivalent  to~$\rho_0$.

Assume $\bm{x}(t)$ and $\bm{x}'(t)$ are two curves parametrized by $\rho_0$-length that differ only in the $m$-th coordinate; denote them by $x_m(t)$ and $x_m'(t)$, respectively.
Note that if $x'_m(t)\le x_m(t)$ for any $t$ and 
the function $x'_m(t)$ is locally $1$-Lipschitz at all $t$ such that $x'_m(t)< x_m(t)$, then 
\[\length_{\rho_1}\bm{x}'\le \length_{\rho_1}\bm{x}.\]
Moreover this inequality is strict if $x'_m(t)< x_m(t)$ for some~$t$.

Fix a curve $\bm{x}(t)$, $t\in[0,\ell]$, parametrized by  $\rho_0$-length.
We can choose $m$ large so that $x_m(t)$ is sufficiently close to $0$ for any~$t$.
In particular, for some values $t$, we have $y_m(t)<x_m(t)$, where
\[y_m(t)=(1-\tfrac t\ell)\cdot x_m(0)
+\tfrac t\ell\cdot x_m(\ell)
-\tfrac 1{100}\cdot \min\{t,\ell-t\}.\]
Consider the curve $\bm{x}'(t)$ as above with
\[x'_m(t)=\min\{x_m(t),y_m(t)\}.\]
Note that $\bm{x}'(t)$ and $\bm{x}(t)$ have the same endpoints, and by the above
\[\length_{\rho_1}\bm{x}'<\length_{\rho_1}\bm{x}.\]
That is, for any curve $\bm{x}(t)$ in $(B,\rho_1)$, we can find a shorter curve $\bm{x}'(t)$ with the same endpoints.
In particular, $(B,\rho_1)$ has no geodesics.

\parbf{\ref{ex:compact+connceted}.}
Choose a sequence of positive numbers $\varepsilon_n\to 0$ and an $\varepsilon_n$-net $N_n$ of $K$ for each $n$.
Assume $N_0$ is a one-point set, so $\eps_0>\diam K$.
Connect each point $x\in N_{k+1}$ to a point $y\in N_{k}$ by a curve of length at most $\eps_k$.

Consider the union $K'$ of all these curves with $K$; observe that $K'$ is compact and path-connected.

\parit{Source:} This problem was suggested by Eugene Bilokopytov \cite{bilokopytov}.

\parbf{\ref{ex:menger}.} Choose two points $x,y\in \spc{X}$;
let $\ell\z=\dist{x}{y}{}$.
Suppose $f\:E\to \spc{X}$ is a distance-preserving map such that $0,\ell\in E\subset [0,\ell]$,
$f(0)=x$, and  $f(\ell)=y$.

Show that we can choose $f$ so that $E$ is maximal;
that is, $f$ cannot be extended to a distance-preserving map on a larger subset of $[0,\ell]$.

Show that there is no open interval $(a,b)$ in the complement of $E$ such that $a,b\in E$.

Apply the completeness of $\spc{X}$ to show that $E$ is closed.
Conclude that $E=[0,\ell]$.

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\parbf{\ref{exercise from BH}.}
Consider the following subset of $\R^2$ equipped with the induced length metric
\[
\spc{X}
=
\bigl((0,1]\times\{0,1\}\bigr)
\cup
\bigl(\{1,\tfrac12,\tfrac13,\dots\}\times[0,1]\bigr).
\]
Note that $\spc{X}$ is locally compact and geodesic.

Its completion $\bar{\spc{X}}$ is isometric to the closure of $\spc{X}$ equipped with the induced length metric;
$\bar{\spc{X}}$ is obtained from $\spc{X}$ by adding two points $p\z=(0,0)$ and $q\z=(0,1)$.

The point $p$ admits no compact neighborhood in $\bar{\spc{X}}$ 
and there is no geodesic connecting $p$ to $q$ in~$\bar{\spc{X}}$.

\parit{Source:} This example is taken from~\cite{bridson-haefliger}.

\parbf{\ref{ex:compact-in-lenght}.}
Let $\spc{X}$ be a compact metric space.
Let us identify $\spc{X}$ with its image in $\Bnd(\spc{X},\RR)$ under the Kuratowsky embedding (Section~\ref{Kuratowsky embedding}). 
Denote by $\spc{K}$ the \textit{linear} convex hull of $\spc{X}$ in the space of bounded functions on $\spc{X}$; 
that is, $x\in \spc{K}$ if and only if $x$ cannot be separated from $\spc{X}$ by a hyperplane.

Since $\spc{X}$ is compact, so is $\spc{K}$.
It remains to observe that $\spc{K}$ is a length space since it is convex.

\parit{Remark.}
Alternatively, one can use the embedding of $\spc{X}$ into its injective hull; see \cite{isbell}.

\parbf{\ref{ex:ultrakatetov}.} 
Let $F=\set{n\in \NN}{f(n)=n}$; we need to show that $\o(F)=1$.

Consider an oriented graph $\Gamma$ with vertex set $\NN\setminus F$ such that $m$ is connected to $n$ if $f(m)=n$.
Show that each connected component of $\Gamma$ has at most one cycle.
Use it to subdivide vertices of $\Gamma$ into three sets $S_1$, $S_2$, and $S_3$ such that $f(S_i)\cap S_i=\emptyset$ for each $i$.

Conclude that $\o(S_1)=\o(S_2)=\o(S_3)=0$ and hence \[\o(F)=\o(\NN\setminus(S_1\cup S_2\cup S_3))=1.\]

\parit{Source:} 
The presented proof was given by Robert Solovay \cite{solovay}, but
the key statement is due to Miroslav Katětov \cite{katetov}.

\parbf{\ref{ex:linear}.}
Choose a nonprincipal ultrafilter $\o$ and set $L(\bm{s})=s_\o$.
It remains to observe that $L$ is linear.

\parit{Remark.}
By this exercise, $\o$ corresponds to a vector in $(\ell^\infty)^*\setminus\ell^1$. 

\parbf{\ref{ex:ultrakatetov+}.}
Use \ref{ex:ultrakatetov}.

\parbf{\ref{ex:Asym(Lob)}}; \ref{SHORT.ex:Asym(Lob):metric-tree}.
Show that there is $\delta>0$ such that sides of any geodesic triangle in $\Lob21$ intersect a disk of radius $\delta$.
Observe that $\Lob2n\z=\tfrac1{\sqrt{n}}\cdot\Lob21$, and use it to show that any geodesic triangle in $\spc{T}$ is a tripod.

\parit{\ref{SHORT.ex:Asym(Lob):homogeneous}.} Observe and use that $\Lob2n$ are homogeneous.

\parit{\ref{SHORT.ex:Asym(Lob):continuum}.} 
Choose $p_1\in \Lob21$, denote by $p_n$ the corresponding point in $\Lob2n\z=\tfrac1{\sqrt{n}}\cdot\Lob21$
Suppose $p_n\to p_\o$ as $n\to\o$; we can assume that $p_\o\in\spc{T}$.
By \ref{SHORT.ex:Asym(Lob):homogeneous}, it is sufficient to show that $p_\o$ has a continuum degree.

Choose distinct geodesics $\gamma_1,\gamma_2\:[0,\infty)\to \Lob21$ that start at a point $p_1$.
Show that the limits of $\gamma_1$ and $\gamma_2$ run in the different connected components of $\spc{T}\setminus \{p_\o\}$.
Since there is a continuum of distinct geodesics starting at $p$,
we get that the degree of $p_\o$ is at least continuum.

On the other hand, the set of sequences of points $q_n\in\Lob2n$  has cardinality continuum.
In particular, the set of points in $\spc{T}$ has cardinality at most continuum.
It follows that the degree of any vertex is at most continuum.

\parit{Remark.}
The properties \ref{SHORT.ex:Asym(Lob):homogeneous} and \ref{SHORT.ex:Asym(Lob):continuum} describe the tree $\spc{T}$ up to isometry \cite{dyubina-polterovich}.
In particular, $\spc{T}$ does not depend on the choice of the ultrafilter.

\parbf{\ref{ex:isom-ultrapower}.}
Show and use that the spaces $\spc{X}^\o$ and $(\spc{X}^\o)^\o$ have discrete metrics and both have cardinality of the continuum.

\parbf{\ref{ex:ultrapower(ultrapower)}.}
Choose a bijection $\iota\:\NN\to \NN\times \NN$.
Given a set $S\subset \NN$, consider the sequence $S_1$, $S_2,\dots$ of subsets in $\NN$ defined by $m\in S_n$ if $(m,n)\z=\iota(k)$ for some $k\in S$.
Set $\o_1(S)=1$ if and only if $\o(S_n)=1$ for $\o$-almost all $n$.
It remains to check that $\o_1$ meets the conditions of the exercise.

\parit{Comment.}
It turns out that $\o_1\ne \o$ for any $\iota$;
see the post of Andreas Blass \cite{blass}.

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\parbf{\ref{ex:notproper-limit}.} Consider the infinite metric $\spc{T}$ tree with unit edges shown
on the diagram. Observe that $\spc{T}$ is proper.

Consider the vertex $v_\o=\lim_{n\to\o}v_n$ in the ultrapower $\spc{T}^\o$.
Observe that $\o$ has an infinite degree.
Conclude that $\spc{T}^\o$ is not locally compact.

\parbf{\ref{ex:nonconvex-limit}.}
Let $\spc{X}_n$ be the square $\{( x,y)\in \R^2, |x|\le 1, |y|\le 1\}$ with the metric induced by the $\ell^n$-norm and let $f_n(x,y)=x$ for all $n$.
Observe that $\spc{X}_\o$ is the square with the metric induced by the $\ell^\infty$-norm where the limit function $f_\o(x,y)=x$ is not concave.

\parbf{\ref{ex:GH-SC}}; \ref{SHORT.ex:GH-SC:circle}.
Suppose $\spc{X}_n\GHto \spc{X}$ and $\spc{X}_n$ are simply connected length metric spaces.
It is sufficient to show that any nontrivial covering map $f\:\tilde{\spc{X}}\to \spc{X}$ corresponds to a nontrivial covering map $f_n\:\tilde{\spc{X}}_n\to \spc{X}_n$ for large $n$.

The latter can be constructed by covering $\spc{X}_n$ by small balls that lie close to sets in $\spc{X}$ evenly covered by $f$, preparing a few copies of these sets and gluing them in the same way as the inverse images of the evenly covered sets in $\spc{X}$ are glued to obtain $\tilde{\spc{X}}$.

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\includegraphics{mppics/pic-2}
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\parit{\ref{SHORT.ex:GH-SC:nonsc-limit}.}
Let $\spc{V}$ be a cone over Hawaiian earrings.
Consider the {}\emph{doubled cone} $\spc{W}$ --- two copies of $\spc{V}$ with  their base points glued (see the diagram).

The space $\spc{W}$ can be equipped with a length metric
(for example, the induced length metric from the shown embedding).

Show that $\spc{V}$ is simply connected, but $\spc{W}$ is not; the latter is a good exercise in topology.

If we delete from the earrings all small circles, then the obtained double cone becomes simply connected and it remains close to $\spc{W}$.
That is, $\spc{W}$ is a Gromov--Hausdorff limit of simply connected spaces.

\parit{Remark.}
Note that from part \ref{SHORT.ex:GH-SC:nonsc-limit}, the limit does not admit a nontrivial covering.
So, if we define the fundamental group as the inverse image of groups of deck transformations for all the coverings of the given space, then one may say that a Gromov--Hausdorff limit of simply connected length spaces is simply connected.

\parbf{\ref{ex:sphere-to-ball},}
\textit{\ref{SHORT.ex:sphere-to-ball:2}.}
Suppose that a metric on $\mathbb{S}^2$ is close to the disk $\DD^2$.
Note that $\mathbb{S}^2$ contains a circle $\gamma$ that is close to the boundary curve of $\DD^2$.
By the Jordan curve theorem, $\gamma$ divides $\mathbb{S}^2$ into two disks, say $D_1$ and $D_2$.

By \ref{ex:GH-SC:circle}, the Gromov--Hausdorff limits of $D_1$ and $D_2$ have to contain the whole $\DD^2$, otherwise the limit would admit a nontrivial covering.

Consider points $p_1\in D_1$ and $p_2\in D_2$ that are close to the center of $\DD^2$.
If $n$ is large, the distance $\dist{p_1}{p_2}{n}$ has to be very small.
On the other hand, any curve from $p_1$ to $p_2$ must cross $\gamma$, so it has length about 2 --- a contradiction.

\parit{\ref{SHORT.ex:sphere-to-ball:3}.}
Make holes in the unit 3-disc, that do not change its topology and do not change its length metric much 
and pass to its doubling in the boundary.

\parit{Source:} The exercise is taken from \cite{burago-burago-ivanov}.

\parbf{\ref{ex:GH-proper-marked}.} Modify proof of \ref{thm:GH-compact}, or apply \ref{thm:ultra-GH:b}.

\parbf{\ref{ex:adjacent-angles}.}
If $\mangle\hinge pxz+\mangle\hinge pyz< \pi$, then by the triangle inequality for angles (\ref{claim:angle-3angle-inq}) we have $\mangle\hinge pxy< \pi$.
The latter implies that $[xy]$ fails to be minimizing near $p$.

\parbf{\ref{ex:tangent-vect=o(t)}.}
By the definition of a right derivative, there is a geodesic $\gamma$ such that both limits 
\[\limsup_{\eps\to0+}\frac{\dist{\alpha(\eps)}{\gamma(\eps)}{\spc{X}}}{\eps}
\quad\text{and}\quad
\limsup_{\eps\to0+}\frac{\dist{\beta(\eps)}{\gamma(\eps)}{\spc{X}}}{\eps}\]
are arbitrarily small.
By the triangle inequality, we get
\[\limsup_{\eps\to0+}\frac{\dist{\alpha(\eps)}{\beta(\eps)}{\spc{X}}}{\eps}=0.\]

\parbf{\ref{ex:both-sided-diff}.}
Apply the definition.

\parbf{\ref{ex:diff}.}
Observe that
\[\speed_t\alpha=|\alpha^+(t)|=|\alpha^-(t)|.\]
Apply Theorem~\ref{thm:speed} to show that
\[\dist{\alpha^+(t)}{\alpha^-(t)}{\T_{\alpha(t)}}=2\cdot\speed_t\alpha.\]


\parbf{\ref{ex:schroeder-foetch}.}
Choose two non-Euclidean norms $\|{*}\|_{\spc{X}}$ and $\|{*}\|_{\spc{Y}}$ on $\RR^{10}$ such that the sum $\|{*}\|_{\spc{X}}\z+\|{*}\|_{\spc{Y}}$ is Euclidean.
See \cite{schroeder-foetch} for more details.

\parbf{\ref{ex:(3+1)-expanding}.} 
Assume $\dist{p}{x^i}{}=\dist{q}{y^i}{}$ for each $i$.
Observe and use that
\[\dist{x^i}{x^j}{}\le\dist{y^i}{y^j}{}
\quad\Longleftrightarrow\quad
\angk\kappa p{x^i}{x^j}\le \angk\kappa q{y^i}{y^j}.\]

\parbf{\ref{ex:(3+1)-expanding}.} Apply the four-point comparison (\ref{df:1+3}).

\parbf{\ref{ex:nongeod-cbb}.}
Modify the induced length metric on the unit sphere in an infinite-dimensional Hilbert space in small neighborhoods of a countable collection of points.
To prove that the obtained space is $\Alex0$, you may need to use the technique from Halbeisen's example (\ref{Halbeisen's example}).

\parbf{\ref{ex:almost.geod}.} Mimic the proof of Theorem~\ref{thm:almost.geod}.

\parbf{\ref{ex:G-delta-not-thru}.}
On the plane, any nonnegatively curved metric having an everywhere dense set of singular points will do the job, where 
by singular point we mean a point having total angle around it strictly smaller than $2\cdot\pi$.

Indeed, if $x_i$ is a singular point, then there is $0<\eps_i<1/20$ such that no geodesic with ends outside of $\oBall(x_i,r)$ can meet the ball $\oBall(x_i,\eps_i\cdot r)$.
The set 
\[\Omega_n=\bigcup_i \oBall(x_i,\tfrac{\eps_i}n)\]
is open and everywhere dense.
Note that $\Omega_n$ may intersect a geodesic  of length $1/n$ only within $\frac 1 {10n}$  of its endpoints.
%V: what was written was incorrect.
%$\tfrac1n$-end of it.
%an endpoint.
The intersection of the $\Omega_n$ is a G-delta dense set that does not intersect the interior of any geodesic.

\parbf{\ref{mink+alex=euclid}.} 
Note that rescaling does not change the space.
Therefore if the space is $\Alex\kappa$, then it is $\Alex{\lambda\cdot\kappa}$ for any $\lambda>0$.
Passing to the limit as $\lambda\to 0$, we may assume that the space is $\Alex0$.

The point-on-side comparison (\ref{point-on-side}) for $p=v$, $x=w$, $y=-w$ and $z=0$ implies that 
\[\|v+w\|^2+\|v-w\|^2\le 2\cdot\|v\|^2+2\cdot\|w\|^2.\]
Applying the comparison for 
$p=v+w$, $x=w-v$, $y=v-w$ and $z=0$ gives the opposite inequality.
That is, the parallelogram identity
\[\|v+w\|^2+\|v-w\|^2= 2\cdot\|v\|^2+2\cdot\|w\|^2\]
holds for any vectors $v$ and $w$.
Whence the statement follows.

\parbf{\ref{ex:cbb-geod-overlap}.}
Apply the hinge comparison (\ref{angle}).

\parbf{\ref{ex:equality-alexlemma}.}
Without loss of generality, we may assume that the points $x,v,w,y$ appear on the geodesic $[xy]$ in that order.
By the point-on-side comparison (\ref{point-on-side}) we have
\begin{align*}
\angk\kappa xyp\le\angk\kappa xwp&\le \angk\kappa xvp,
\\
\angk\kappa ywp&\ge\angk\kappa yvp\ge\angk\kappa yxp.
\end{align*}
Therefore
\begin{align*}\angk\kappa xyp<\angk\kappa xwp
\quad&\Longrightarrow\quad
\angk\kappa xyp<\angk\kappa xvp,
\\
\angk\kappa yxp<\angk\kappa ywp
\quad&\Longleftarrow\quad
\angk\kappa yxp<\angk\kappa yvp.
\end{align*}

By Alexandrov's lemma (\ref{lem:alex}), we have
\begin{align*}
\angk\kappa xyp<\angk\kappa xvp
\quad&\Longleftrightarrow\quad
\angk\kappa yxp<\angk\kappa yvp,
\\
\angk\kappa xyp<\angk\kappa xwp
\quad&\Longleftrightarrow\quad
\angk\kappa yxp<\angk\kappa ywp.
\end{align*}
Hence the statement follows.


\parbf{\ref{ex:urysohn}.}
See the construction of Urysohn's space \cite[3.11$\tfrac{3}{2}_+$]{gromov-MS} or \cite{petrunin2020pure}.

\parbf{\ref{ex:lebedeva-petrunin}.}
Read \cite{lebedeva-petrunin}.

\parbf{\ref{ex:fat-triangle}.}
Apply the angle-sidelength  monotonicity (\ref{cor:monoton}) twice. 

\parbf{\ref{ex:busemann}.}
The first part follows from the angle-sidelength  monotonicity (\ref{cor:monoton}).
An example for the second part can be found among metrics on $\RR^2$ induced by a norm. (Compare to Exercise~\ref{mink+alex=euclid}.)

\parit{Remark.} This exercise is inspired by Busemann's definition \cite{busemann-CBA}.

\parbf{\ref{ex:busemann-CBB} and \ref{ex:busemann-CBA}}; \ref{SHORT.ex:busemann-CBB:a}.
By the function comparison definitions of $\Alex{0}$ space (\ref{comp-kappa}), for any $p\in \spc{L}$ and $\eps>0$ the function $\distfun{p}{}{}$ is $\eps$-concave everywhere sufficiently far from $p$.
Applying the definition of Busemann function, we get the result.

The $\CAT0$ case is analogous; we have to apply (\ref{function-comp}) and use $\eps$-convexity.

\parit{\ref{SHORT.ex:busemann-CBB:b}.}
By the definition of Busemann function (see  \ref{prop:busemann}),
\begin{align*}
\exp(\bus_\gamma) 
&=  \lim_{t\to \infty} \exp(\distfun{{\gamma (t)}}{}{} - t)=
\\
&= \lim_{t\to \infty} \left[\exp (\distfun{\gamma (t)}{}{} -t)
+\exp(-\distfun{\gamma (t)}-t)\right]=
\\
&=  \lim_{t\to \infty} \left(2\cdot \exp(-t)\cdot \cosh \circ\distfun{\gamma (t)}{}{}\right).
\end{align*}

By the function comparison definitions of $\CAT\kappa$ space (\ref{function-comp}) or $\Alex{\kappa}$ space (\ref{comp-kappa}),  for any $p\in \spc{U}$ the function $f=\cosh \circ\distfun{p}{}{}$ satisfies $f''+\kappa \cdot f\ge 1$ (respectively, $f''+\kappa \cdot f\le 1$).
The result follows.

\parbf{\ref{ex:noncomplete-globalization}.}
Read \cite{petrunin:globalization}.

\parbf{\ref{ex:fixed-point}.} If $\diam(\spc{L}/G)>\tfrac\pi2$, then for some $x\in \spc{L}$ we have
\[\sup \set{\distfun{G\cdot x}(y)}{y\in \spc{L}}
>
\tfrac\pi2.\]
Use comparison to show that there is a unique point $y^{*}$ that lies at maximal distance from the orbit $G\cdot x$.
Observe that $y^{*}$ is a fixed point.

\parbf{\ref{ex:kleiner}.}
Assume there are 4 such points $x_1,x_2,x_3,x_4$.
Since the space $\spc{L}$ is $\Alex{1}$, it is also $\Alex{0}$.
By the angle comparison, the sum of the angles in any geodesic triangle in an $\Alex{0}$ space is $\ge \pi$.
Therefore the average of the $\mangle\hinge{x_i}{x_j}{x_k}$ is  larger than $\tfrac\pi3$.
On the other hand, since each $x_i$ has space of directions $\le\tfrac12\cdot\mathbb{S}^n$ and the perimeter of any triangle in $\tfrac12\cdot\mathbb{S}^n$ is at most $\pi$, the average of $\mangle\hinge{x_i}{x_j}{x_k}$ is at most $\tfrac\pi3$ --- a contradiction.

\parit{Source:} Based on the main idea in \cite{hsiang-kleiner}.

\parbf{\ref{ex:ccat-(3+1)}.}
Suppose that 
\[\angk\kappa {x^0}{x^1}{x^2}+\angk\kappa {x^0}{x^2}{x^3}<\angk\kappa {x^0}{x^1}{x^3}.\]
Show that
\[\angk\kappa {x^2}{x^0}{x^1}+\angk\kappa {x^2}{x^1}{x^3}+\angk\kappa {x^2}{x^3}{x^0}>2\cdot\pi.\]
Conclude that one can take $p=x^2$.

\parbf{\ref{ex:sba-2+2-short}.}
Modify the configuration in \ref{def:2+2-reformulated}.

\parbf{\ref{ex:berg-nikolaev}.}
Read \cite{sato};
the original proof \cite{berg-nikolaev} is harder to follow.

An example for the second part of the problem can be found among 4-point metric spaces.
It is sufficient to take four vertices of a generic convex quadrangle and increase one of its diagonals slightly;
it will still satisfy the inequality for all relabeling but will fail to meet \ref{def:2+2-reformulated}.

\parbf{\ref{ex:CAT-mnfld=>ext.geod}.}
Suppose that a geodesic $[px]$ is not extendable beyond $x$.
We may assume that $\dist{p}{x}{}<\varpi\kappa$;
otherwise move $p$ along the geodesic toward  $x$.

By the uniqueness of geodesics (\ref{thm:cat-unique}), any point $y$ in a neighborhood $\Omega\ni x$ is connected to $p$ by a unique geodesic path; denote it by $\gamma_y$.
Note that $h_t(y)=\gamma_y(t)$ defines a homotopy, called the  \index{geodesic!homotopy}\emph{geodesic homotopy}, between the identity map of $\Omega$ and the constant map with value~$p$.

Since $[px]$ is not extendable, $x\notin h_t(\Omega)$ for any $t<1$.
In particular, the local homology groups vanish at $x$ --- a contradiction.

\parbf{\ref{ex:complete-space-of-dir}.}
Choose a sequence of directions $\xi_n$ at $p$;
by $\gamma_n\:\RR\to \spc{U}$ the corresponding local geodesics.
Since the space $\spc{U}$ is locally compact, we may pass to a converging subsequence of $(\gamma_n)$; its limit is a local geodesic by Corollary~\ref{cor:loc-geod-are-min}.
Denote the limit by  $\gamma_\infty$ and its direction by $\xi_\infty$.
By comparison, $\xi_\infty$ is a limit of $(\xi_n)$.

\parbf{\ref{mink+CAT=euclid}.} Follow the solution in the \ref{mink+alex=euclid}, reversing all the inequalities.

\parbf{\ref{ex:convexity-CAT0}.} 
It is sufficient to show that if $v$ and $y$ are midpoints of geodesics $[uw]$ and $[xz]$ in $\spc{U}$, then
\[\dist{v}{y}{}\le \tfrac12\cdot(\dist{u}{x}{}+\dist{w}{z}{}).\]

\begin{wrapfigure}{r}{35 mm}
\vskip-5mm
\centering
\includegraphics{mppics/pic-10}
\end{wrapfigure}

Denote by $p$ the midpoint of $[uz]$.
Applying the angle-sidelength  monotonicity (\ref{cor:monoton-cba}) twice, we have
\[\dist{v}{p}{}\le \tfrac12\cdot\dist{w}{z}{}.\]
Similarly we have
\[\dist{y}{p}{}\le \tfrac12\cdot\dist{u}{x}{}.\]
It remains to add these two inequalities and apply the triangle inequality.

\parit{Remark.}
This inequality also follows directly from the majorization theorem (\ref{thm:major}).

\parbf{\ref{ex:equality-for-thin}.}
The only-if part is evident.
Use \ref{thm:defs_of_cat} to show that 
\ref{SHORT.ex:equality-for-thin:side-side}$\Rightarrow$\ref{SHORT.ex:equality-for-thin:vertex-base}$\Rightarrow$\ref{SHORT.ex:equality-for-thin:angle}.
By \ref{thm:defs_of_cat}, condition \ref{SHORT.ex:equality-for-thin:angle} implies that the natural map is distance-preserving on the sides $[\tilde x\tilde y]$ and $[\tilde x\tilde z]$.
Applying it again, we have that condition \ref{SHORT.ex:equality-for-thin:angle} holds for all permutations of the labels $x,y,z$.
Whence the natural map is distance-preserving on all three sides.

\parit{Remark.}
These conditions imply that the natural map can be extended to a distance-preserving map to the solid model triangle.
In fact the image of the line-of-sight map (\ref{def:sight}) is isometric to the model triangle.

\parbf{\ref{ex:busemann-CBA}.}
See the solution of Exercise~\ref{ex:busemann-CBB}.

\parbf{\ref{ex:patchwork}}; \ref{SHORT.ex:patchwork:proper}.
Suppose that $x_n\to x_\infty$, $y_n\to y_\infty$ as $n\to\infty$,
but $[x_ny_n]$ does not converge to $[x_\infty y_\infty]$.
Since the space is proper, we can pass to a subsequence such that $[x_ny_n]$ converges to another geodesic.
That is, we have at least two geodesics between $x_\infty$ and $y_\infty$.

\parit{\ref{SHORT.ex:patchwork:complete}.}
Let $\Delta_n$ be a sequence of solid spherical triangles 
with angle $\tfrac\pi4$ and adjacent sides $\pi-\tfrac1n$.
Let us glue each $\Delta_n$ to $[0,\pi]$ along an isometry of one of the longer sides.
It remains to show that the obtained space $\spc{X}$ is a needed example.

\parit{Source:}
The example \ref{SHORT.ex:patchwork:complete} is taken from \cite[Chapter I, Exercise 3.14]{bridson-haefliger}.

\parbf{\ref{ex:two-rays}.}
Subdivide $Q$ into a a half-plane $A$ bounded by the extension of $\gamma_1$ and the remaining solid angle $B$; it has angle measure measure $\pi-\alpha$.
First glue $B$ along $\gamma_2$, and then glue $A$.
Each time apply the Reshetnyak gluing theorem (\ref{thm:gluing}), to show that the obltained space is $\CAT0$.

\parbf{\ref{ex:reshetnyak-doubling}.}
Suppose that $A$ is not convex.
Then there is a geodesic $[xy]$ with ends in $A$ that does not lie in $A$ completely.
Note that $[xy]$ can be lifted to two different geodesics with the same ends  in the doubling, and apply uniqueness of geodesics (\ref{cor:cat-unique}).

\parbf{\ref{ex:glue-spherical-suspension}.}
Since $K$ is $\pi$-convex, it is $\CAT1$.
By \ref{thm:warp-curv-bound:cat}, the spherical suspension $\Susp K$ is $\CAT1$ as well.
Let us glue $\Susp K$ to $\spc{U}$  along $K$;
according to the Reshetnyak gluing theorem, the resulting space, say $\spc{U}'$, is $\CAT1$.

Consider the geodesic path $\gamma\:[0,1]$ from $p$ to a pole of the suspension in $\spc{U}'$.
Set $K_t=\spc{U}\cap\cBall[\gamma(t),\tfrac\pi2]$.
By \ref{cor:convex-balls}, $K_t$ is $\pi$-convex for any $t$;
monotonicity and continuity of the family should be evident.

\parit{Source:}
This construction was used in \cite{lytchak-petrunin-2020}.
Applying it together with Sharafutdinov retraction leads to another solution of Exercise~\ref{ex:short-retraction-CBA(1)}.

\parbf{\ref{ex:AUB}.}
Apply the Reshetnyak gluing theorem, or its reformulation \ref{thm:gluing2-reformulated}.


\parbf{\ref{ex:fenchel}.} By \ref{thm:major}, there is a majorization $F\:D\to\spc{U}$ of the polygonal line $\beta$.
Show and use that $D$ is a convex plane polygon and its external angles cannot exceed the corresponding external angle of $\beta$.


\parbf{\ref{ex:FM}.} This exercise generalizes the so-called Fáry–Milnor theorem.
An elementary proof is given in the first author and Richard Bishop \cite{alexander-bishop:fm};
another proof is given by Stephan Stadler \cite{stadler}.

\parbf{\ref{ex:isometric-majorization}.}
\textit{(Easier way.)} 
Let 
$(t,s)\mapsto \gamma_t(s)$ be the line-of-sight map 
for $\alpha$ with respect to $\alpha(0)$,
and 
$(t,s)\mapsto \tilde \gamma_t(s)$ be the line-of-sight map 
for $\tilde \alpha$ with respect to $\tilde \alpha(0)$.
Consider the map  $F\:\Conv\tilde \alpha\to \spc{U}$ such that 
$F\:\tilde \gamma_t(s)\mapsto \gamma_t(s)$.

Show that $F$ majorizes $\alpha$
and conclude that $F$ is distance-preserving.

\parit{(Harder way.)}
Prove and apply the following statement together with the Majorization theorem.
\begin{itemize}
\item Let $\alpha$ and $\beta$ be two convex curves in $\Lob2\kappa$.
Assume 
\[\length \alpha=\length\beta<2\cdot\varpi\kappa\]
and there is a short bijecction $f\:\alpha\to\beta$.
Then $f$ is an isometry.
\end{itemize}

\parbf{\ref{ex:bishop}.}
Suppose that points $p,x,q,y$ appear on the curve in that cyclic order.
Assume that the geodesics $[pq]$ and $[xy]$ do not intersect.
Use the argument in the proof of the majorization theorem (\ref{thm:major}) to show that in this case there are nonequivalent majorization maps.

Now we can assume that pairs of geodesics $[pq]$ and $[xy]$ intersect for all choices of points $p,x,q,y$ on the curve in that cyclic order.
Show that in this case the convex hull $K$ of the curve is isometric to a convex figure.

Note that the composition of a majorization map and closest point projection to $K$ is a majorization.
Show and use that the boundary of a convex figure in the plane admits a unique majorization up to equivalence.

\parit{Remark.}
A typical rectifiable closed curve in a $\CAT0$ space can be majorized by more than one convex figure.
There are two exceptions: (1) the majorization map is distance-preserving, and (2) the curve is geodesic triangle.
It is expected that there are no other exceptions;
this question was asked by Richard Bishop in a private conversation.

\parbf{\ref{ex:square}.}
Show that quadrangle $[x^1x^2x^3x^4]$ is majorized by the solid quadrangle $[\tilde x^1\tilde x^2\tilde x^3\tilde x^4]$.
Further show that the majorization is isometric;
argue as in \ref{ex:equality-alexlemma}.

\parbf{\ref{ex:cover-branching-along-2-lines}.}
If $\ell$ and $m$ do not intersect, then the double cover $\spc{X}$ is not simply connected.
In particular, by the Hadamard--Cartan theorem, $\spc{X}$ is not $\CAT0$.

If $\ell$ and $m$ intersect, then $\spc{X}$ is a cone over a double cover $\Sigma$ of $\mathbb{S}^2$ branching at two pairs $(x,y)$ and $(v,w)$ of antipodal points.
Suppose $\dist{x}{v}{\mathbb{S}^2}=\ell<\tfrac\pi2$.
Note that the inverse image of $[xv]_{\mathbb{S}^2}$ is a closed geodesic of length $4\cdot\ell<2\cdot\pi$.
Therefore, by the generalized Hadamard--Cartan theorem, $\Sigma$ is not $\CAT1$. Hence $\spc{X}$ is not $\CAT0$ by Theorem \ref{thm:warp-curv-bound:cat}.

\parbf{\ref{ex:branching}.}
Let us do the second part first.
Assume $A$ has nonempty interior. 
Note that the space $\tilde{\spc{U}}$ is simply connected and locally isometric to the doubling $\spc{W}$ of $\spc{U}$ in $A$;
that is, any point in $\tilde{\spc{U}}$ has a neighborhood 
%\Omega$ 
that is isometric to a neighborhood of a point in $\spc{W}$.

By the Reshetnyak gluing theorem (\ref{thm:gluing}), $\spc{W}$ is $\CAT0$.
Therefore $\tilde{\spc{U}}$ is locally $\CAT0$;
it remains to apply the Hadamard--Cartan theorem (\ref{thm:hadamard-cartan}).

Let us come back to the general case.
The above argument can be applied to a closed $\eps$-neighborhood of $A$.
After that we need to pass to a limit as $\eps\to 0$.

The first part of the problem follows since a geodesic is a convex set.

\begin{wrapfigure}{r}{25 mm}
\vskip-4mm
\centering
\includegraphics{mppics/pic-11}
\end{wrapfigure}

\parbf{\ref{ex:geod-circle}.}
By the globalization theorem there is  a nontrivial homotopy class in the set of all closed curves of length $<2\cdot\varpi\kappa$.
Consider a shortest noncontractible closed curve $\gamma$ in  $\spc{K}$;
note that such a curve exists.
Indeed, let $L$ be the infimum of lengths of all noncontractible closed curves in $\spc{X}$.
Compactness and local contractibility of $\spc{X}$ imply that any two sufficiently close closed curves in $\spc{X}$ are homotopic.
Then choosing a sequence of unit speed noncontractible curves whose lengths converge to $L$, an Arzel\'{a}--Ascoli type of argument shows that these curves subconverge to a noncontractible curve of length~$L$.

Assume that $\gamma$ is not a geodesic circle,
that is,  there are two points $p$ and $q$ on $\gamma$ such that the distance $\dist{p}{q}{}$ 
is shorter than the lengths of the arcs, say $\alpha_1$ and $\alpha_2$, of $\gamma$ from $p$ to~$q$.
Consider the products, say $\gamma_1$ and $\gamma_2$,
of $[qp]$ with $\alpha_1$ and~$\alpha_2$.
Then $\length\gamma_1, \length\gamma_2<\length \gamma$.

Therefore, showing that $\gamma_1$ or $\gamma_2$ is noncontractible in the set of curves of length $<2\cdot\varpi\kappa$, will lead to a contradiction.
The case $\kappa\le 0$ is trivial.
For the case $\kappa> 0$, argue as in \ref{thm:hadamard-cartan-gen}.

\parit{Remark.}
The statement of the exercise fails if the requirement that $\spc{X}$ be compact is replaced by the assumption that it is proper.
For example, the surface of revolution of the graph of $y=e^x$ around the $x$-axis is locally $\CAT0$ but has no closed geodesics.

\parbf{\ref{ex:closest-point-projection}.}
Let $p\mapsto\bar p$ denote the closest-point projection to $K$.
We need to show that $\dist{\bar p}{\bar q}{}\le\dist{ p}{ q}{}$ for any $p,q\in \spc{U}$.

Assume $p\ne \bar p\ne \bar q\ne q$.
Note that in this case $\mangle\hinge {\bar p}{p}{\bar q}\ge \tfrac\pi2$ and $\mangle\hinge {\bar q}{q}{\bar p}\ge \tfrac\pi2$.
Otherwise a point on the geodesic $[\bar p\bar q]$ would be closer to $p$ or to $q$ than $\bar p$ or $\bar q$, respectively.
The latter is impossible since $K$ is convex and therefore $[\bar p\bar q]\subset K$.

Applying the arm lemma (\ref{lem:arm}), we get the statement.

The cases $p= \bar p\ne \bar q\ne q$ and $p\ne \bar p\ne \bar q= q$ can be done similarly.
The rest of the cases are trivial.

\parbf{\ref{ex:short-retraction-CBA(1)}.}
A more transparent, but less elementary solution via gradient flow is given by Alexander Lytchak and the third author~\cite{lytchak-petrunin-2020}.

\medskip

Without loss of generality, we may assume that $p\in K$.

If $\distfun{K}{x}{}\ge\pi$, then set $\map[2](x)=p$.

Otherwise, if $\distfun{K}{x}{}<\pi$, by the closest-point projection lemma (\ref{lem:closest point}), 
there is a unique point $x^*\in K$ that minimizes distance to $x$;
that is, $\dist{x^*}{x}{}=\distfun{K}{x}{}$.
Let us define $\ell_x$, $\phi_x$ and $\psi_x$ using the following identities:
\begin{align*}
\ell_x&=\dist{p}{x^*}{},
\\
\phi_x&=\tfrac\pi2-\dist[{{}}]{x^*}{x}{},
\\
\sin\psi_x&=\sin\phi_x\cdot\sin\ell_x, 
\quad 0\le \psi_x\le \tfrac\pi2.
\intertext{Let}
\map[2](x)&=\geod_{[px^*]}(\psi_x).
\end{align*}

Note that $\map[2]$ is a retraction to $K$; 
that is,
$\map[2](x)\in K$ for any $x\in \spc{U}$
and 
$\map[2](a)=a$ for any $a\in K$.

Let us show that $\map[2]$ is short.
Given $x,y\in\oBall(K,\tfrac\pi2)$, let
\begin{align*}
x'&=\map[2](x)
&
y'&=\map[2](y)
\\
r&=\dist{x}{y}{}
&
r'&=\dist{x'}{y'}{}
\\
d&=\dist{x^*}{y^*}{}
&
\alpha&=\angk1{p}{x^*}{y^*}.
\end{align*}

Note that 
\[\cos r\le 
\cos\phi_x\cdot\cos\phi_y
-
\cos d\cdot\sin\phi_x\cdot\sin\phi_y.
\eqlbl{eq:cos(r)}\]

Indeed, if $x,y\notin K$,
then 
$\mangle\hinge{x^*}{x}{y*}, 
\mangle\hinge{y^*}{y}{x*}
\ge 
\tfrac\pi2$
and inequality~\ref{eq:cos(r)} follows from the arm lemma (\ref{lem:arm}).
If $x\in K$ and $y\notin K$, we obtain \ref{eq:cos(r)} by the angle comparison (\ref{cat-hinge}) 
since $\mangle\hinge{y^*}{y}{x*}\ge \tfrac\pi2$.
In the same way, \ref{eq:cos(r)} is proved 
if $x\notin K$ and $y\in K$.
Finally, if $x,y\in K$, then $\phi_x=\phi_y=\tfrac\pi2$ and $r=d$;
that is, the inequality trivially holds.

Further note that
\[\cos\alpha
=
\frac{\cos d-\cos \ell_x\cdot\cos\ell_y}{\sin\ell_x\cdot\sin\ell_y}.\]
Applying the angle-sidelength  monotonicity (\ref{cor:monoton-cba}), we have
\begin{align*}
\cos r'&\ge
\cos\psi_x\cdot\cos\psi_y
-
\cos \alpha \cdot\sin\psi_x\cdot\sin\psi_y=
\\
&=
\cos\psi_x\cdot\cos\psi_y
-(\cos d-\cos \ell_x\cdot\cos\ell_y)\cdot\sin\phi_x\cdot\sin\phi_y\ge
\\
&\ge \cos\psi_x\cdot\cos\psi_y
-\cos d\cdot\sin\phi_x\cdot\sin\phi_y.
\end{align*}


Note that 
$\psi_x\le \phi_x$
and
$\psi_y\le \phi_y$;
in particular,
\[
\cos\phi_x\cdot\cos\phi_y\le \cos\psi_x\cdot\cos\psi_y.
\]
Hence 
\[\cos r'\ge \cos r;\]
that is, the restriction $\map[2]|_{\oBall(K,\tfrac\pi2)}$ is short.
Clearly $\map[2]$ is continuous. 
Since the complement of $\oBall(K,\tfrac\pi2)$ is mapped to $p$,
 $\map[2]$ is short; that is,
\[r'\le r \eqlbl{eq:cos=<cos}\]
for any $x,y\in\spc{U}$.

If we have equality in \ref{eq:cos=<cos}
then 
\[\cos\ell_x\cdot\cos\ell_y\cdot\sin\phi_x\cdot\sin\phi_y=0.\]
If $K\subset \oBall(p,\tfrac\pi2)$, then $\ell_x,\ell_y<\tfrac\pi2$, 
which implies that $x\in K$ or $y\in K$.
Without loss of generality, we may assume that $x\in K$.

It remains to show that if $y\notin K$, 
then the inequality~\ref{eq:cos=<cos}
is strict.
If $\distfun{K}{y}{}\ge\tfrac\pi2$, then \ref{eq:cos=<cos} holds since 
the left-hand side is $<\tfrac\pi2$
while the right-hand side is $\ge \tfrac\pi2$.
If $\distfun{K}{y}{}<\tfrac\pi2$, then $\phi_y>0$. Clearly $\psi_y<\phi_y$,
hence inequality~\ref{eq:cos=<cos} is strict.
\qeds

Below you will find a geometric way to think about the given construction; 
it is close to the construction in the proof of Kirszbraun's theorem (\ref{thm:kirsz+}).

\parit{Geometric interpretation of the map $\map[2]$.}
Let $\mathring{\spc{U}}=\Cone \spc{U}$, and 
denote by $\mathring{K}$ the subcone of $\mathring{\spc{U}}$ spanned by $K$.
The space $\spc{U}$ can be naturally identified with the unit sphere in $\mathring{\spc{U}}$, 
that is, the set 
\[\set{z\in \mathring{\spc{U}}}{|z|=1}.\]

According to \ref{thm:warp-curv-bound:cat}, $\mathring{\spc{U}}$ is $\CAT0$.
Note that $\mathring{K}$ forms a convex closed subset of $\mathring{\spc{U}}$.
According to \ref{lem:closest point}, for any point $x$ there is a unique point $\hat x\in \mathring{K}$
that minimizes the distance to $x$,
that is, $\dist{\hat x}{x}{}=\distfun{K}{x}{}$.
(If $|\hat x|\ne0$, then in the notation above we have
$x^*=\tfrac1{|\hat x|}\cdot\hat x$.)

Consider the half-line $t\mapsto t\cdot p$ in  $\mathring{\spc{U}}$.
By comparison, 
for given $s\in \mathring{\spc{U}}$
the geodesics $\geod_{[s\ t\cdot p]}$ converge as $t\to\infty$ to a half-line, 
say $\alpha_s\:[0,\infty)\to \mathring{\spc{U}}$.



Note that if $|x|=1$, then $|\hat x|\le 1$.
By assumption, for any $a\in K$ the function $t\mapsto |\alpha_a(t)|$ is monotonically increasing.
Therefore there is a unique value $t_x\ge 0$ such that
$|\alpha_{\hat x}(t_x)|=1$.
Define $\map[2]\:\spc{U}\to K$
 by 
\[\map[2](x)=\alpha_{\hat x}(t_x).\]

\parbf{\ref{ex:cats-cradle}.}
Prove that the angle comparison (\ref{cat-hinge}) holds.

\parbf{\ref{ex:Hadamard--Cartan}.}
Mimic the proof of the Hadamard--Cartan theorem.

\parbf{\ref{ex:CBB+CBA}.}
Note that it is sufficient to show that any finite set of points $x^1,\dots,x^n\in\spc{X}$ lies in an isometric copy of a Euclidean polyhedron.

Observe that $\spc{X}$ is $\Alex0$ and $\CAT0$ at the same time.
Show that there is a unique point $p$ that minimizes the sum $\dist{p}{x^1}{}+\dots+\dist{p}{x^n}{}$.
Note that the vectors $v^i=\ddir{p}{x^i}$ lie in a linear subspace of $\T_p$.
Moreover if $K$ is the convex hull of $v_i$, then the origin of $\T_p$ lies in the interior of $K$ relative to its affine hull.
Finally observe that the exponential map is defined on all of $K$ and is distance-preserving.
The statement follows since the exponential map sends $v^i\mapsto x^i$ for each $i$.

\parbf{\ref{ex:5-point-CBA=>CBB}.}
The answers are $s\le \sqrt3$ and $s\le 2$, respectively.

Let us start with the $\CAT0$ case.
The upper bound $s\le \sqrt3$ follows from (2+2)-point comparison.
The Euclidean space works as an example if $s$ is smaller than the large diagonal of the double pyramid with unit side (that is, if $s\le 2\cdot\sqrt{2/3}$).
Otherwise it can be embedded into a product of the real line with a two-dimensional cone.

For the $\Alex0$ case, the needed space can be constructed by doubling  a polyhedron $K\subset\EE^3$ in its boundary.
The obtained space is $\Alex0$ by \ref{thm:poly-CBB};
the same follows from Perelman's doubling theorem \cite{perelman:spaces2}.
We assume that the points correspond to vertices of a regular tetrahedron with 3 vertices on the boundary of $K$ and one  in its interior; this point corresponds to a pair of points in the doubling at distance $s$ from each other.

\parit{Remark.}
The $\CAT0$ case also follows from \cite{toyoda,lebedeva-petrunin:toyoda}.

\parbf{\ref{ex:cbb-wald}.}
Choose a quadruple of points $p,q,r,s$. 
Suppose that it admits a distance-preserving embedding into some $\Lob2{\Kappa}$ for some $\Kappa\ge \kappa$.
Then 
\[\angk\Kappa p{q}{r}
+\angk\Kappa p{r}{s}
+\angk\Kappa p{a}{q}\le 2\cdot\pi.\]
Applying monotonicity of the function $\kappa\mapsto\angk\kappa p{q}{r}$ (\ref{k-decrease}) shows  that
\[\angk\kappa p{q}{r}
+\angk\kappa p{r}{s}
+\angk\kappa p{s}{q}\le 2\cdot\pi.\]
Since the quadruple $p,q,r,s$ is arbitrary, the if part follows.

Now let us prove the only-if part.
Denote by $\sigma$ the exact upper bound on values $\Kappa\ge \kappa$ such that all model triangles with the vertices $p,q,r,s$ are defined.

Recall that $\angk{\Kappa+} p{q}{r}$ denotes extended angle (\ref{def:extended-angle}).
Observe that if 
\[\angk{\Kappa+} p{q}{r}
+\angk{\Kappa+} p{r}{s}
+\angk{\Kappa+} p{s}{q}= 2\cdot\pi\eqlbl{eq:Kappa3}\]
for some $\sigma\ge \Kappa\ge \kappa$, then the quadruple admits a distance-preserving embedding into $\Lob2\Kappa$.

Observe that the left-hand side of \ref{eq:Kappa3} is continuous in $\Kappa$.
Since $\spc{L}$ is $\Alex\kappa$, for $\Kappa=\kappa$ the left-hand side cannot exceed $2\cdot \pi$.
Therefore it remains smaller than $2\cdot\pi$ for all $\sigma\ge \Kappa\ge \kappa$;
moreover the same holds for all permutations of the labels $p,q,r,s$.

Note that we can assume the perimeter of the triple $q,r,s$ is $2\cdot\varpi{\sigma}$, and use this and the overlap lemma (\ref{lem:extend-overlap}) to arrive at a contradiction.

According to our definition, the real line is $\Alex\kappa$ for any $\kappa\in\RR$,
but it does not satisfy the property for $\kappa>0$. 
The condition $\kappa\le 0$ was used just once to ensure that the $\kappa$-model triangles with the vertices $p,q,r,s$ are defined.
One can assume instead that perimeters of all triangles in $\spc{L}$ are at most $2\cdot\varpi\kappa$.
This condition holds for all complete length $\Alex\kappa$ spaces of dimension at least 2; see \ref{diam-k>0}.

\parbf{\ref{ex:sturm}.}
Let $\tilde p,\tilde x_1,\dots,\tilde x_n$ be the array in $\EE^n$ provided by the (1+\textit{n})-point comparison (\ref{thm:pos-config}).
We may assume that $\tilde p$ is the origin of $\EE^n$.

Consider an $n{\times}n$-matrix $\tilde M$ with components 
\[\tilde m_{i,j}=\tfrac12\cdot(\dist[2]{\tilde x_i}{\tilde p}{}+\dist[2]{\tilde x_j}{\tilde p}{}-\dist[2]{\tilde x_i}{\tilde x_j}{}).\]
Note that $\tilde m_{i,j}=\langle\tilde x_i,\tilde x_j\rangle$.
It follows that $\tilde M=A\cdot A^\top$ for an $n{\times}n$-matrix $A$ that defines a linear transformation sending the standard basis to the array $\tilde x_1,\dots,\tilde x_n$.
Therefore
\[\bm{s}\cdot \tilde M\cdot \bm{s}^\top=|A^\top\cdot \bm{s}^\top|^2 \ge 0\]
for any vector $\bm{s}$.
Further show and use that
\[\bm{s}\cdot M\cdot \bm{s}^\top\ge \bm{s}\cdot \tilde M\cdot \bm{s}^\top\]
for any vector $\bm{s}=(s_1,\dots,s_n)$ with nonnegative components.

\parbf{\ref{6-point-comparison}.} Apply the (5+1)-point comparison (\ref{thm:pos-config}).

\parbf{\ref{ex:(3+1)-nonsufficient}.}
It is sufficient to construct a metric on the set of points $\{p$, $x^1$, $x^2$, $x^3$, $x^4\}$ that does not satisfy (1+4)-point comparison but does satisfy all (1+3)-point comparisons.
To do this, set  $x^i$ to be  the vertices of a regular tetrahedron in $\EE^3$. Suppose $p$ is its center and reduce the distances $\dist{p}{x^i}{}$ slightly.

\parit{Remark.}
There are examples of 6-point metric spaces that satisfy all (1+5)-point comparisons, but do not admit embedding into a complete length $\Alex{0}$ space \cite{lebedeva-petrunin-zolotov}.


\parbf{\ref{ex:strut+embedding}.}
By the (1+\textit{n})-point comparison (\ref{thm:pos-config}), there is a point array $\tilde p,\tilde a^0,\dots,\tilde a^m\in \Lob{m+1}\kappa$ such that
\[\dist{\tilde p}{\tilde a^i}{}=\dist{p}{a^i}{}\quad \text{and}\quad \dist{\tilde a^i}{\tilde a^j}{}\ge\dist{a^i}{a^j}{}\]
for all $i$ and $j$.

For each $i$, set 
$\tilde \xi^i=\dir{\tilde p}{\tilde a^i}\in\mathbb{S}^m=\Sigma_{\tilde p}(\Lob{m+1}\kappa)$.
Note that 
\[\dist{\tilde \xi^i}{\tilde \xi^j}{\mathbb{S}^m}\ge \angk\kappa{p}{ a^i}{ a^j}>\tfrac\pi2.\]

Consider two matrices $S$ and $\tilde S$ with components
$s_{i,j}=\langle\tilde \xi^i,\xi^j\rangle$
and
$\tilde s_{i,j}=\cos[\angk\kappa{p}{a^i}{a^j}]$.
By construction, $S\ge 0$; that is $\bm{v}\cdot S\cdot \bm{v}^\top\ge 0$ for any vector $\bm{v}$.

Observe that it is sufficient to show that $\tilde S\ge 0$.
The latter follows since $s_{i,j}\le \tilde s_{i,j}\le 0$ if $i\ne j$ and
$s_{i,j}= \tilde s_{i,j}=1$ if $i=j$.

\parbf{\ref{ex:flat-in-CAT}.}
Set $\tilde Q=\Conv\{\tilde x^0,\tilde x^1,\dots,\tilde x^\kay\}$.
By Kirszbraun's theorem, the map $\tilde x^i\mapsto x^i$ can be extended to a short map $F\:\tilde Q\to\spc{L}$;
it remains to show that the map $F$ is distance-preserving.

Consider the logarithm map $G\:x\mapsto \ddir{x_0}x$; note that $G$ is short.
Observe that the composition $G\circ F$ is distance-preserving.
Therefore $F$ is distance-preserving;
in particular we can take $Q=F(\tilde Q)$.

\parbf{\ref{ex:flat-in-CBB}.}
Consider vectors $v^i=\ddir{x^0}{x^i}\in\T_{x^0}$.
Show that all the $v^i$ lie in a linear subspace of $\T_{x^0}$ and that $x^i\mapsto v^i$ is distance-preserving.
It follows that we can identify the convex hull $K$ of  the $v^i$ with the convex hull of  the $\tilde x^i$.

Note that the gradient exponential map $\gexp_{x_0}$ maps $v^i$ to $x^i$.
By assumption, 
\[\dist{v^i}{v^j}{}=\dist{x^i}{x^j}{}\eqlbl{eq:vv=xx}\]
for all $i$ and $j$.
By \ref{thm:prop-gexp}, $\gexp_{x_0}$ is a short map.
By \ref{eq:vv=xx}, $\gexp_{x_0}$ cannot be strictly short at a pair of points in $K$.
That is, $\gexp_{x_0}$ is distance-preserving on $K$.

\parbf{\ref{CBA-n-point}.}
Apply \ref{thm:kirsz} for each of the following maps:
\begin{itemize}
\item $f_0\:\tilde x\mapsto x$, $\tilde p^1\mapsto p^1$, $\tilde q^1\mapsto q^1$;
\item $f_i\:\tilde p^i\mapsto p^i$, $\tilde p^{i+1}\mapsto p^{i+1}$, $\tilde q^i\mapsto q^i$, $\tilde q^{i+1}\mapsto q^{i+1}$ for $1\le i<n$;
\item $f_n\:\tilde y\mapsto y$, $\tilde p^n\mapsto p^n$, $\tilde q^n\mapsto q^n$.
\end{itemize}
Denote by $F_i$ the short extension of $f_i$.
Observe and use that $F_{i-1}(\tilde z_i)\z=F_{i}(\tilde z_i)$ for each $i$.

\parbf{\ref{ex:petrunin-stadler}.} Consider the space $\spc{Y}^{\spc{X}}$ of all maps $\spc{X}\to \spc{Y}$ equipped with the product topology.

Denote by $\mathfrak{S}_F$ the set of maps $h\in \spc{Y}^\spc{X}$ such that the restriction $h|_F$  is short and agrees with $f$ in $F\cap A$.
Note that the sets $\mathfrak{S}_F\subset \spc{Y}^\spc{X}$ are closed and any finite intersection of these sets is nonempty.

According to Tikhonov's theorem, $\spc{Y}^{\spc{X}}$ is compact.
By the finite intersection property, the intersection $\bigcap_F\mathfrak{S}_F$ for all finite sets $F\subset X$ is nonempty.
Hence the statement follows.

\parit{Source:} This statement appears in \cite{petrunin-stadler}; it is an analogous of the finite+one lemma (\ref{lem:kirsz-neg:new}).

\parbf{\ref{ex:isbell}.}
The Kuratowsky embedding is a distance-preserving map of $\spc{X}$ into the space of bounded functions $\spc{X}$ equipped with the metric induced by the sup-norm (Section~\ref{Kuratowsky embedding}).
It remains to show that the latter space is injective.

The second part of the exercise is a classical result of John Isbell \cite{isbell} which was rediscovered several times after him; for more on the subject see lecture notes of the third author \cite{petrunin2020pure}.

\parbf{\ref{ex:warp=<}.}
It is sufficient to show that the natural map $\spc{B}\warp{g}\spc{F}\to \spc{B}\warp{f}\spc{F}$ is short.
The latter follows from the fiber-independence theorem (\ref{thm:fiber-independence}).

\parbf{\ref{ex:convexity-in-cone}.}
Show and use that any geodesic path in $\Cone^\kappa\spc{F}$ projects to a reparametrized geodesic in $\spc{F}$ of length less than $\pi$.

\parbf{\ref{ex:spherical-join}.}
By \ref{thm:warp-curv-bound:cbb:a}, the space $\spc{U}$, $\spc{V}$, or $\spc{U}\star\spc{V}$ is $\Alex1$ if and only if $\Cone\spc{U}$, $\Cone\spc{V}$, or $\Cone(\spc{U}\star\spc{V})=\Cone\spc{U}\times\Cone\spc{V}$ is $\Alex0$, respectively.

By \ref{thm:warp-curv-bound:cbb:S}, the space $\spc{U}$, $\spc{V}$, or $\spc{U}\star\spc{V}$ is $\CAT1$ if and only if $\Cone\spc{U}$, $\Cone\spc{V}$, or $\Cone(\spc{U}\star\spc{V})=\Cone\spc{U}\times\Cone\spc{V}$ is $\CAT0$, respectively.

It remains to show that the product of two spaces is $\Alex0$ or $\CAT0$ if and only if each space is $\Alex0$ or $\CAT0$, respectively.

\parbf{\ref{ex:metric tree}.}
Apply Reshetnyak gluing theorem (\ref{thm:gluing}) several times.

\parbf{\ref{ex:poly-unique-geodesic}.}
Assume $\spc{P}$ is not $\CAT0$.
Then by \ref{thm:PL-CAT}, a link $\Sigma$ of some simplex contains a closed local geodesic $\alpha$ with length $4\cdot\ell<2\cdot\pi$.
We can assume that $\Sigma$ has minimal possible dimension;
then by \ref{thm:PL-CAT}, $\Sigma$ is locally $\CAT1$.

Divide $\alpha$ into two equal arcs $\alpha_1$ and $\alpha_2$.

Assume $\alpha_1$ and $\alpha_2$ are length-minimizing, and 
parametrize them by $[-\ell,\ell]$.
Fix a small $\delta>0$ and 
consider the two curves in $\Cone\Sigma$ given in polar coordinates by 
\[\gamma_i(t)=(\alpha_i(\arctan \tfrac t\delta),\sqrt{\delta^2+t^2}).\]
Show that the curves $\gamma_1$ and $\gamma_2$ are geodesics in $\Cone\Sigma$ having common endpoints.

Observe that a small neighborhood of the tip of $\Cone\Sigma$ admits a distance-preserving embedding into~$\spc{P}$.
Hence we can construct two geodesics $\gamma_1$ and $\gamma_2$ in $\spc{P}$ with common endpoints.

It remains to consider the case where $\alpha_1$ (and therefore $\alpha_2$) is not length-minimizing.

Pass to a maximal length-minimizing arc $\bar\alpha_1$ of $\alpha_1$.
Since $\Sigma$ is locally $\CAT1$, by the no-conjugate-point theorem (\ref{thm:no-conj-pt}) 
there is another geodesic $\bar\alpha_2$ in $\Sigma_p$ that shares endpoints with $\bar\alpha_1$.
It remains to repeat the above construction for the pair $\bar\alpha_1$, $\bar\alpha_2$.

\parit{Remark.}
By \ref{thm:cat-unique} the converse holds as well.
This problem was suggested by Dmitri Burago.

\parbf{\ref{ex:polyKk}.} Apply \ref{thm:tan-is}, \ref{thm:poly-CBB}, and \ref{thm:PL-CAT}.

\parbf{\ref{ex:barycenric-flag}.}
Observe and use that (1) in the barycentric subdivision every vertex corresponds to a simplex of the original triangulation,
and (2) a simplex of the subdivision corresponds to a decreasing sequence of simplexes in the original triangulation. 

\parit{Remark.}
The second statement, \textit{any finite  simplicial complex is homeomorphic to a compact length $\CAT1$ space}, is due to Valerii Berestovskii \cite{berestovskii}.

\parbf{\ref{ex:obtuce-flag}.}
Use induction on the dimension  to prove that if in a spherical simplex $\triangle$ every edge is at least $\tfrac\pi2$, then 
all dihedral angles of $\triangle$ are at least~$\tfrac\pi2$.

The rest of the proof goes along the same lines as the proof of the flag condition (\ref{thm:flag}).
The only difference is that a geodesic may spend time \textit{at least} $\pi$ on each visit to $\Star_v$.

\parit{Remark.}
It is not sufficient to assume only that all the dihedral angles of the simplexes are at least~$\tfrac\pi2$. 
Indeed, the two-dimensional sphere with the interior of a small rhombus removed is a spherical polyhedral space glued from four triangles with angles at least~$\tfrac\pi2$.
On the other hand, the boundary of the rhombus is a closed local geodesic in this space and has length less than $2\cdot\pi$.
Therefore the space cannot be $\CAT1$.

\parbf{\ref{ex:short+commuting}.}
Observe that if we glue two copies of spaces along $A_i$, then the copies of $A_j$ for some $j\ne i$ form a convex subset in the glued space.
Use this and the Reshetnyak gluing theorem (\ref{thm:gluing}) $n$ times, once for each label of the edges.

\parbf{\ref{ex:space-of-trees}.}
The space $\spc{T}_n$ has a natural cone structure whose vertex is the  completely degenerate tree --- all its edges have zero length.

Note that the space $\Sigma$
over which the cone is taken comes naturally with a triangulation 
by right-angled spherical simplexes.
Each simplex corresponds to the combinatorics of a possibly degenerate tree.

Note that the link of any simplex of this triangulation satisfies the no-triangle condition.
Indeed, fix a simplex $\triangle$ of the complex;
suppose it is described by a possibly degenerate topological tree $t$.
A triangle in the link of  $\triangle$ can be described by three ways to resolve a degeneracy of $t$ by adding one edge, where 
(1) any pair of these resolutions can be done simultaneously, but (2) all three cannot be done simultaneously.
Direct inspection shows that this is impossible.

By Proposition~\ref{prop:no-trig}, our complex is flag.
It remains to apply the flag condition (\ref{thm:flag}) and \ref{thm:warp-curv-bound:cbb:a}.

\parbf{\ref{ex:cubical-complex}.}
Apply the flag condition (\ref{thm:flag}) and Theorem~\ref{thm:warp-curv-bound:cbb:S}.

\parbf{\ref{ex:norays}.}
Consider a cube in the $\ell^2$-space defined by $|x_i|\le 1$.

\parbf{\ref{ex:d_q dist_p(v)=-<dri p q, v>-CAT} and \ref{ex:d_q dist_p(v)=-<dri p q, v>}.}
Apply the strong angle lemmas
(\ref{lem:strong-angle-cba}
and \ref{lem:strong-angle}).

\parbf{\ref{ex:grad(dist)}.} Apply \ref{ex:d_q dist_p(v)=-<dri p q, v>-CAT}.

\parbf{\ref{ex:|grad|=1}.} Apply \ref{thm:almost.geod}.

\parbf{\ref{ex:df(v)=<grad f,v>}.}
Since $\alpha$ is Lipschitz, so is $f\circ\alpha$.
By the standard Rademacher theorem, the derivative $(f\circ\alpha)'$ is defined almost everywhere.
In particular, 
\[(\dd_{\alpha(t)}f)(\alpha^+(t))+(\dd_{\alpha(t)}f)(\alpha^-(t))\ae0.\]

Further, by the extended Rademacher theorem (more precisely, its 1-dimensional case; see Proposition~\ref{prop:Rademacher-dim=1}),
we have 
\[\alpha^+(t)+\alpha^-(t)\ae0.\]
In particular,
\[\<\nabla_{\alpha(t)}f,\alpha^+(t)\>+\<\nabla_{\alpha(t)}f,\alpha^-(t)\>
\ae0.\]

Finally, by the definition of gradient, we have 
\[\<\nabla_{\alpha(t)}f,\alpha^\pm(t)\>\ge (\dd_{\alpha(t)}f)(\alpha^\pm(t)).\]
Hence the result follows.

\parbf{\ref{ex:d dist(grad)<0}.}
Let us pass to the ultrapower $\spc{L}^\o\supset\spc{L}$.
Argue as in \ref{ex:d_q dist_p(v)=-<dri p q, v>} to show that there is a geodesic $[pb]_{\spc{L}^\o}$ such that 
\[\dd_p\distfun b{}(\dir pa)=-\langle\dir pb,\dir pa\rangle.\]

It follows that
\begin{align*}(\dd_p\distfun{a}{}{})(\nabla_p\distfun{b}{}{})&\le-\langle\dir pa,\nabla_p\distfun{b}{}{} \rangle\le
\\
&\le -\dd_p\distfun{b}{}{}(\dir pa)=
\\
&=\langle\dir pb, \dir pa\rangle=
\\
&= \cos\mangle\hinge p a b _{\spc{L}^\o}\le 
\\
&\le \cos\angk\kappa p a b.
\end{align*}

\parbf{\ref{ex:fixed-point-CAT}.}
Choose an orbit $\{p_1,\dots,p_n\}$;
observe that its barycenter with equal masses is a fixed point.  

\parbf{\ref{ex:helly}.}
Let $k$ be the maximal integer such that any $k$ subsets have a common points.
By assumption $k\ge m+1$.

Suppose $k<n$.
We can assume that $n=k+1$ and $K_1\cap\dots\cap K_{n}=\emptyset$.
Choose a point
\[p_i\in \bigcap_{j\ne i} K_j\]
for each $i$.
Argue as in \ref{lem:nondeg-test-with-balls} to show that the simplex with vertices $p_1,\dots,p_n$ is nondegenerate and arrive at a contradiction.

\parbf{\ref{ex:compact-dimension-cbb}.}
Suppose that $\spc{L}$ is infinite-dimensional.
Denote by $\Omega_m\subset \spc{L}$ the set of all points $p$ with $\rank_p\ge m$.
Evidently $\Omega_1\supset \Omega_2\supset\dots$, and $\Omega_m$ is open for each $m$.

By \ref{LinDim+}, each $\Omega_m$ is dense in $\spc{L}$.
Hence there is a G-delta dense set of points $p\in\spc{L}$ such that $\rank_p=\infty$.
It follows that $\Sigma_p$ is not compact.

\parit{Source:} It was suggested by Alexander Lytchak.

\parbf{\ref{ex:sharafutdinov}.}
Choose a finite sequence $t_0<\dots<t_n$.
Denote by $\Phi_{t_i}$ the composition of the closest-point projections to $K_{t_0},\dots, K_{t_i}$.
Pass to a limit of the $\Phi_{t_i}$ as the sequence becomes denser in the parameter interval. 
Show and use that the limit $\phi_t$ does not depend on the choice of the sequences. 


\parbf{\ref{ex:elf-contracting}.}
Let $\ell(t)=\dist{\alpha(t)}{\alpha(t_3)}{}$.
Note that 
\[\ell'(t)\le -\langle \nabla_{\alpha(t)}f,\dir{\alpha(t)}{\alpha(t_3)}\rangle.\]

Observe that the function $t\mapsto f\circ\alpha(t)$ is nondecreasing;
in particular, $f(\alpha(t_1))\le f(\alpha(t_2))\le f(\alpha(t_3))$.
Therefore 
\begin{align*}\langle \nabla_{\alpha(t)}f,\dir{\alpha(t)}{\alpha(t_3)}\rangle&\ge\dd_{\alpha(t)}f(\dir{\alpha(t)}{\alpha(t_3)})\ge 0
\end{align*}
for any $t\in[t_1,t_2]$.
Therefore $\ell'\le 0$ for any $t\in[t_1,t_2]$.
Hence the statement.

\parbf{\ref{ex:grad-curve-condition}.} 
Without loss of generality, we may assume that $(f\circ\alpha)'(t)>0$ for any $t$.
Let $\hat\alpha$ be the arclength reparametrization of $\alpha$.
Note that 
\[(f\circ\hat\alpha)'(s)\ge |\nabla_{\hat\alpha(s)}f|\]
almost everywhere.
Therefore, by Theorem~\ref{thm:grad-like-2nd-def}, $\hat\alpha$ is a gradient-like curve.
It remains to apply Lemma~\ref{lem:grad--grad-like}.

\parbf{\ref{ex:grad-curve-analitic}.}
Use \ref{ex:grad-curve-condition} to prove the only-if part.

To prove the if part, set $h(z)=\tfrac12\cdot\dist[2]{x}{z}{}$.
If $\alpha$ is an $f$-gradient curve, then 
\begin{align*}
(h\circ\alpha)^+&\ge \dist{\alpha(t)}{x}{}\cdot\langle\dir{\alpha(t)}{x},\nabla_{\alpha(t)}f\rangle\ge
\\
&\ge \dist{\alpha(t)}{x}{}\cdot\dd_{\alpha(t)}f(\dir{\alpha(t)}{x})\ge 
\\
&\ge f(x)-f\circ\alpha(t).
\end{align*}
It remains to integrate the inequality and observe that $f\circ\alpha$ is nondecreasing.

\parbf{\ref{ex:geodesic}.}
Consider $(x,\kappa)$- and $(z,\kappa)$-radial curves that start at $y$,
and observe that they form a geodesic from $x$ to $z$.
(Compare to Exercise~\ref{ex:flat-in-CBB}.)

\parbf{\ref{ex:gexp}.} Set $q=p+v$ and $q'=\gexp_pv$. 
By radial comparison, $\dist{q'}{x}{}\le \dist{q}{x}{}$ for any $x\in \spc{L}$.
If $q\in \spc{L}$, this implies that $q=q'$.
Otherwise note that $q'$ lies on the boundary line of $\spc{L}$, and $\proj(q)$ is the only point on this line that satisfies the inequality.

\parbf{\ref{ex:inv-gexp}.}
By the angle comparison,
$|\nabla_x\distfun p|\ge-\cos \angk\kappa xpq$.

Choose a $(p,\kappa)$-radial curve $\alpha$ that starts at $p$.
Observe that 
\[(\distfun p\circ \alpha)^+(t)\ge-|\alpha^+(t)|\cdot \cos \angk\kappa {\alpha(t)}p q\]
and
\[(\distfun q\circ\alpha)^+(t)\ge -|\alpha^+(t)|.\]
Therefore $t\mapsto\angk\kappa q{\alpha(t)}p$  is nondecreasing, hence the result.