\section*{Introduction}
\subsection*{Lipschitz extension problem}
Let $X$, $Y$ be metric spaces.
Consider $X'\subset X$ and a Lipschitz map $f:X'\rightarrow Y$.
Can we extend $f$ to $F:X\rightarrow Y$ with the same constant $\mathrm{Lip}(F)=\mathrm{Lip}(f)$? 
Failing that, can we bound the loss?
This potential ``loss'' can be encapsulated in a function $\mathcal{L}_{X,Y}$:
$$\begin{array}{rlcl} \mathcal{L}_{X,Y}: & \mathbb{R}^+ & \longrightarrow & \quad \mathbb{R}^+ \\ & C & \longmapsto &\displaystyle  
\sup_{\substack{X'\subset X \\ f:X'\rightarrow Y \\ \mathrm{Lip}(f)\leq C}} \; 
\inf_{\substack{F:X\rightarrow Y \\ F|_{X'}=f}} \; 
\mathrm{Lip}(F).\end{array}$$
For example, maps to~$\mathbb{R}$, or more generally to a metric tree $T$, can always be extended without loss~\cite{m, las}: $\mathcal{L}_{X, \mathbb{R}}(C)=\mathcal{L}_{X, T}(C)=C$ for all $C\geq 0$.
Kirszbraun~\cite{k} proved that $\mathcal{L}_{X, Y}(C)=C$ when $X,Y$ are Euclidean spaces.

Recall that a Hadamard manifold is a complete, simply connected Riemannian manifold of nonpositive sectional curvature.
Lang and Schr\"oder~\cite{las}, extending work of Valentine~\cite{v} for the constant-curvature case, proved:
\begin{theoremA} \cite{las}
Let $\kappa_0, \kappa'_0<0$ be constants.
If $X,Y$ are Hadamard manifolds with $\kappa_X \geq \kappa_0$ and $\kappa_Y \leq \kappa'_0$, then $\mathcal{L}_{X,Y}(C)=C$ for all $C\geq \sqrt{\kappa_0/\kappa'_0}$.
\label{lasch}
\end{theoremA}
\subsubsection*{Main result}
Up to scaling, we may and always will assume $\kappa_0=\kappa'_0=-1$.
In that case, the above theorem also gives: $\mathcal{L}_{X,Y}(C)\leq 1$ when $C\leq 1$. 
The goal of this note is to prove the following refinement:
\begin{theorem} \label{main}
For any $C<1$, $K\leq -1$ and $m \in \mathbb{N}$, there exists $C'<1$ such that 
for any Hadamard manifolds $X,Y$ of dimension $\leq m$ 
satisfying $\kappa_X\geq -1 \geq \kappa_Y \geq K$, 
one has $\mathcal{L}_{X,Y}(C)\leq C'$.
\end{theorem}
The $X=Y=\mathbb{H}^2$ case was conjectured in~\cite[App.\ C]{guk}, which put forward a strategy when $X'$ has bounded diameter.

\subsection*{About the method}
Our proof is based on the template of Lang and Schr\"oder's proof of Theorem~A, which we will recall in~\S\ref{prolasch} (slightly simplified, as~\cite{las} is set in the context of Alexandrov spaces).
The extra ingredients, which extend and uniformize arguments of~\cite{guk}, are based on the notion that under negative curvature, both in the small-scale limit (Euclidean geometry) and large-scale limit (real trees), loss-less extension is known to hold.
Thus, loss ($\mathcal{L}_{X,Y}(C)>C$) is in a sense a medium-range phenomenon, and can be controlled using a form of compactness and covering arguments.

When extending $f:X'\rightarrow Y$ to a single point $\xi \in X\smallsetminus X'$, we will see in~\S\ref{prolasch} that there is usually a natural ``optimal'' image $F(\xi)$, relative to the set $X'$ where the map is already defined. 
Given a second point $\xi'$, we can then assign it an optimal image relative to $X'\cup \{\xi\}$, then pass to a third point $\xi''$ and so on, studying the loss incurred at each step.
One difficulty, which could cause the losses to pile up, is that the notion of ``optimal'', being relative to $X'\cup\{\xi, \xi', \dots\}$, changes as we go.

However, as pointed out in~\cite{las}, this difficulty \emph{disappears} when $Y$ is a metric tree: then, taking each $\xi\in X\smallsetminus X'$ to its optimal image (relative to $X'$ only) yields a globally Lipschitz map, with no loss.
This key feature, together with the fact that the curvature bounds force $Y$ coarsely to behave somewhat like a tree at large distances, is what allows us to prove Theorem~\ref{main}.
To patch together maps defined on different regions of $X$, we will use a standard interpolation procedure described in~\S\ref{interpol}.

\subsubsection*{Plan}
Section~\S\ref{prolasch} recalls the proof of Theorem~A; Section~\S\ref{secmain} proves Theorem~\ref{main}. Section~\S\ref{seconj} indulges in some speculation.

\subsection*{Notation}
Distances in metric spaces are all denoted $d$.

\noindent The open ball centered at $\xi$, of radius $r$, is written $\mathbb{B}_\xi(r)$. For a ball of unspecified center, we sometimes write $\mathbb{B}(r)$.

\noindent Given a point $\xi$ in a Hadamard manifold $X$, we write $\exp_\xi: \mathrm{T}_\xi(X) \rightarrow X$ the exponential map, and $\log_\xi$ its inverse.

\noindent Given $x,z\in X\smallsetminus \{\xi\}$, the notation $\widehat{x\xi z} \in [0,\pi]$ then refers to the angle between vectors $\log_\xi(x)$ and $\log_\xi(z)$, for the Euclidean metric on $\mathrm{T}_\xi(X)$.

\noindent The volume measure on $X$ is written $\mathrm{Vol}_X$.

\subsection*{Acknowledgements}
My interest for the Lipschitz extension problem in negative curvature started with Fanny Kassel and our work~\cite{guk}, which tackled an equivariant version. I am grateful to her and to Urs Lang for discussions related to this material.

\section{Proof of Theorem~A}\label{prolasch}
To build loss-less extensions, it is enough to do it one point $\xi\in X$ at a time: indeed, we can then repeat for a dense sequence $(\xi_n)_{n\in \mathbb{N}}$ of $X$, and pass to all of $X$ by continuity. 

Let $X'\subset X$ and $f:X'\rightarrow Y$ be $C$-Lipschitz with 
$$C\geq 1,$$
where $X,Y$ are Hadamard manifolds subject to curvature bounds $\kappa_X \geq -1 \geq \kappa_Y$.
We can restrict attention to $X'$ compact, nonempty. Consider $\xi \in X\smallsetminus X'$: the function defined by
$$\begin{array}{rcl} \varphi_\xi:  Y & \longrightarrow & \mathbb{R}^+  \\ 
    y  & \longmapsto & \displaystyle \max_{x\in X'} \, \frac{d(y,f(x))}{d(\xi, x)}
\end{array}$$
is proper and convex on $Y$, hence achieves a minimum 
\begin{equation}\label{cxi} C_\xi:=\min \varphi_\xi = \varphi_\xi (\eta) \end{equation}
at some $\eta\in Y$ (in fact unique). 
We can think of $\eta$ as an ``optimal candidate for $F(\xi)$'': Theorem~A will follow if we can prove 
$$C_\xi \leq C.$$ 
If $C_\xi \leq 1$ we are done. If $C_\xi\geq 1$, define the compact set 
\begin{equation}\label{xxi} X_\xi:=\Big \{x\in X', \frac{d(\eta, f(x))}{d(\xi, x)}=C_\xi \Big \}. \end{equation}

The exponential of any linear hyperplane $V \subset \mathrm{T}_\eta Y$ separates $Y$ into two half-spaces, each of which contains points of $f(X_\xi)$ in its closure: if not, we could push $\eta$ towards $f(X_\xi)$ (perpendicularly to $V$) to reduce $\varphi_\xi(\eta)$, contradicting minimality. 
Hence, $\eta$ belongs to the convex hull of some points $y_i=f(x_i)$, $1\leq i \leq n$ where $x_i \in X_\xi$:
\begin{equation} \label{convhull} \sum_{i=0}^n \lambda_i \log_\eta (y_i) = 0_\eta \in \mathrm{T}_\eta Y \end{equation}
for some reals $\lambda_i>0$.
Since $x_i\in X_\xi$, the lengths $\ell_i:=d(\xi,x_i)= \Vert \log_\xi (x_i) \Vert$ satisfy $C_\xi \ell_i = d(\eta, y_i)= \Vert \log_\eta (y_i) \Vert$. 
We can then write
\begin{align*} 0 &\leq \Big \Vert C_\xi \sum_{i=0}^n \lambda_i \log_\xi (x_i) \Big \Vert^2 - \Big \Vert \sum_{i=0}^n \lambda_i \log_\eta (y_i) \Big \Vert^2 \\
&= C_\xi^2 \sum_{i,j} \lambda_i \lambda_j \,  \ell_i \ell_j \, \big( \cos \widehat{x_i \xi x_j} - \cos \widehat{y_i \eta y_j} \big )
\end{align*}
hence at least one summand with $i\neq j$ is $\geq 0$, which happens if and only if $\widehat{x_i \xi x_j} \leq \widehat{y_i \eta y_j}$. 
Hence, up to reindexing, we may assume 
\begin{equation} \theta := \widehat{x_1 \xi x_2} \leq \widehat{y_1\eta y_2} =:\theta'. \label{cot} \end{equation}


Let $\mathcal{D}_\theta(\ell, \ell')$ denote the distance, in the hyperbolic plane $\mathbb{H}^2$, between the far ends of two segments of lengths $\ell, \ell'$ starting from a common vertex, an angle $\theta$ apart.
A~well-known trigonometric formula gives explicitly
\begin{equation} \label{trigo}
\mathcal{D}_\theta(\ell, \ell') = \mathrm{Arccosh} (\cosh \ell \, \cosh \ell' - \sinh \ell \, \sinh \ell' \, \cos \theta) 
\end{equation}
but we will mostly use the following facts: the function $\mathcal{D}_\theta$ is convex in its two arguments, vanishes at $(0,0)$, and depends monotonically on $\theta$.
The Cartan-Alexandrov-Toponogov or CAT($-1$) comparison inequalities, whose interesting history is recounted in~\cite{cat}, say that
\begin{equation} \label{CAT}
\begin{array}{llll}
\mathrm{a.} & d(x_1, x_2) & \leq & \mathcal{D}_{\theta}(\ell_1, \ell_2) \\ 
\mathrm{b.} & d(y_1, y_2) & \geq & \mathcal{D}_{\theta'}(C_\xi \ell_1, C_\xi \ell_2) 
\end{array} \end{equation}
due to the curvature bounds $\kappa_X \geq -1 \geq \kappa_Y$.
Therefore,

\begin{align} 
 C\, d(x_1, x_2) & \geq  d(y_1, y_2) & \text{(Lipschitz bound)} \notag \\
& \geq \mathcal{D}_{\theta'}(C_\xi\ell_1, C_\xi\ell_2) & \text{by \eqref{CAT}}&\mathrm{.b} \notag \\
& \geq \mathcal{D}_{\theta}(C_\xi\ell_1, C_\xi\ell_2) & \text{by \eqref{cot}} \notag \\ 
& \geq  C_\xi \, \mathcal{D}_{\theta}(\ell_1, \ell_2) & \label{conv} \\
& \geq  C_\xi \, d(x_1, x_2). & \text{by \eqref{CAT}}&\mathrm{.a} \notag 
\end{align}
where \eqref{conv} uses convexity of $\mathcal{D}_\theta$ and $C_\xi\geq 1$. Hence $C_\xi\leq C$ as desired, 
proving Theorem~A. \qed

\section{Proof of Theorem~\ref{main}} \label{secmain}

\subsection{One-point extension}
We start by bounding the loss for extensions to a single point.

\begin{lemma} \label{1point}
For any $C<1$ there exists $C^*<1$ such that for any Hadamard manifolds $X,Y$ satisfying $\kappa_X \geq -1 \geq \kappa_Y$, any $X'\subset X$ and any $\xi\in X\smallsetminus X'$, every $C$-Lipschitz map $f:X'\rightarrow Y$ has a $C^*$-Lipschitz extension to $X'\sqcup \{\xi\}$.
\end{lemma}
\begin{proof}
Take $f, C, \xi$ as in the statement and define $C_\xi\geq 0$ (as well as $\eta \in Y$, $X_\xi \subset X'$, $y_i=f(x_i)\in f(X_\xi)$ and $\ell_i=d(\xi, x_i)$) as in the previous proof. 
Theorem~A gives $C_\xi\leq 1$: let us bound $C_\xi$ away from $1$ in terms of $C$ alone.

Let $\Delta>0$ be such that
\begin{equation} \label{addloss} \mathcal{D}_{\pi/2}(\ell, \ell') \geq \ell + \ell' - \Delta ~\text{ for all }~ \ell, \ell'\geq 0 \end{equation}
(using~\eqref{trigo} one can show $\Delta=\log 2$ works).
Let $r>0$ be large enough that 
\begin{equation} \label{C'} \widehat C:=C+{\Delta}/{r}<1. \end{equation}
We distinguish two cases.

$\bullet$ If $\ell_i \geq r$ for some index $i$, we use~\eqref{convhull} to find $j\neq i$ such that 
\begin{equation} \label{obtus} \theta':= \widehat{y_i \eta y_j} \geq \pi/2 \end{equation} 
and write:
\begin{align} 
C\, d(x_i, x_j) & \geq  d(y_i, y_j) & \text{(Lipschitz bound)} \notag \\
& \geq \mathcal{D}_{\theta'}(C_\xi\ell_i, C_\xi\ell_j) & \text{by \eqref{CAT}}&\mathrm{.b} \notag \\
& \geq C_\xi (\ell_i + \ell_j) - \Delta & \text{by \eqref{addloss}--\eqref{obtus}} \notag \end{align}
hence $$C_\xi\leq
\widehat C$$ 
by \eqref{C'}, due to the triangle inequality $d(x_i, x_j) \leq \ell_i+\ell_j$ and $\ell_i \geq r$.


$\bullet$ If no such index $i$ exists, then we define $x_1, x_2\in X_\xi$ and $\theta\leq \theta' \in [0,\pi]$ as in the proof of Theorem~A and write, similar to~\eqref{conv}:
\begin{align}
C\, d(x_1, x_2) & \geq  d(y_1, y_2) & \text{(Lipschitz bound)} \notag \\
& \geq \mathcal{D}_{\theta'}(C_\xi\ell_1, C_\xi\ell_2) & \text{by \eqref{CAT}}&\mathrm{.b} \notag \\
& \geq \mathcal{D}_{\theta}(C_\xi\ell_1, C_\xi\ell_2) & \text{by \eqref{cot}} \notag \\
& \geq C'_\xi \, \mathcal{D}_{\theta}(\ell_1, \ell_2) & \text{(\emph{see \eqref{newc} below})}\label{pattes} \\ 
& \geq C'_\xi \, d(x_1, x_2) & \text{by \eqref{CAT}}&\mathrm{.a} \notag
\end{align}
where we use the new constant 
\begin{equation} \label{newc} C'_\xi:= \frac{\sinh (C_\xi r)}{ \sinh (r)}. \end{equation}
Indeed, for a basepoint $o\in \mathbb{H}^2$, the differential of the exponential map $\mathrm{exp}_o:(\mathbb{R}^2,0) \rightarrow (\mathbb{H}^2,o)$ at 
a point of the circle $\partial \mathbb{B}_0(\lambda)$
has principal values~$1$ radially and $\sinh (\lambda)$ along the circle --- this can be checked by differentiating~\eqref{trigo} near $(\ell, \ell', \theta)=(\lambda, \lambda, 0)$. 
It follows that the radial map 
$$ \begin{array}{rrcl} H: & (\mathbb{H}^2,o) & \longrightarrow & (\mathbb{H}^2,o) \\
& x & \longmapsto & \exp_{o} \big (C_\xi \log_o (x) \big ), \end{array} $$
defining a homothety of ratio $C_\xi$ on each line through $o$, satisfies
$$\mathrm{Lip}\left ( \left . H^{-1} \right |_{\mathbb{B}_o(C_\xi r)} \right ) = \frac{\sinh (r)}{\sinh(C_\xi r)}= \frac{1}{C'_\xi}$$
which means that step \eqref{pattes} holds (using $\ell_1, \ell_2\leq r$). 
Therefore, $C'_\xi \leq C$. 
Substituting in~\eqref{newc}, we find
\begin{equation} \label{arcsinh} C_\xi \leq \frac{\mathrm{Arcsinh}(C\sinh (r))}{r}<1. \end{equation}

In either case, we have bounded the Lipschitz constant $\mathrm{max}\{C,C_\xi\}$ (for the one-point extension $\xi \mapsto \eta$ of $f$) uniformly away from $1$.
\end{proof}

\subsection{Averaging maps} \label{interpol}
In curvature $\leq 0$, convex interpolation behaves well with respect to Lipschitz constants.
Namely, given $f_0, f_1 : X \rightarrow Y$, let $(f_t(x))_{t\in [0,1]}$ be the constant-speed parametrization of the geodesic segment $[f_0(x), f_1(x)]$, for all $x\in X$.
The ``barycenter'' maps $f_t:X\rightarrow Y$ thereby defined satisfy:
\emph{if $f_1$ agrees with $f_0$ on $X' \subset X$ then so does $f_t$.}

Moreover, for all $x,x'\in X$, if $(y_t)_{t\in [0,1]}$ denotes the constant-speed para\-metrization of the segment $[f_0 (x), f_1(x')]$, then
\begin{align*}
 d(f_t(x),f_t(x')) & \leq d(f_t(x),y_t) + d(y_t,f_t(x')) \\
 & \leq t\,  d(f_1(x),f_1(x')) + (1-t)\, d(f_0(x),f_0(x'))
\end{align*}
by CAT(0) comparison inequalities. It follows that 
$$ \mathrm{Lip}(f_t) \leq t\, \mathrm{Lip}(f_1) + (1-t)\, \mathrm{Lip}(f_0).$$
We will simply use the \emph{notation} $$f_t =: t\, f_1 + (1-t)\, f_0.$$
We can also iterate the construction above, to define barycenters of $N$ maps: given maps $(f_i)_{i\geq 1}$, the maps $F_N=\sum_{i=1}^N \frac{1}{N} f_i$, defined inductively on $N$ by 
$F_N : = \frac{1}{N} f_N + \frac{N-1}{N} \big ( \sum_{i=1}^{N-1} \frac{1}{N-1} f_i \big )$, inductively satisfy for any $Z\subset X$:
\begin{equation} \label{moyenne} 
\mathrm{Lip}(F_N|_Z) \leq \sum_{i=1}^N \frac{1}{N} \mathrm{Lip}(f_i|_Z).
\end{equation}
When $N\geq 3$ this construction is not robust under permutation of the $f_i$; note however that symmetric constructions do exist~\cite{las}, which also satisfy a weakened form of associativity~\cite{guk}.

\subsection{Extensions to the whole space}
We now prove Theorem~\ref{main}. 
Let $C<1$, $K\leq -1$, $m\in \mathbb{N}$ and Hadamard manifolds $X,Y$ be as in the theorem, and $C^*\in[C,1)$ be given by Lemma~\ref{1point}.

Let $f:X'\rightarrow Y$ be a $C$-Lipschitz map, where $X'\subset X$.
Again, we may assume $X'$ is compact.
By Lemma~\ref{1point}, we may consider a family of $C^*$-Lipschitz extensions $(f_\xi^*)_{\xi\in X}$ to $X'\cup\{\xi\}$, taking $\xi$ to its optimal candidate image. 
We do allow $\xi \in X'$, in which case $f_\xi^*=f$.
Small balls in $X$ and $Y$ are uniformly $(1+o(1))$-bi-Lipschitz to Euclidean balls, by the curvature bounds $0\geq \kappa_X, \kappa_Y \geq K$ 
(in fact CAT-type inequalities~\eqref{CAT} show that this $o(1)$ tolerance is quadratic in the size of the balls). 
By composition, loss-less extension in Euclidean geometry~\cite{k} implies that there exists $\varepsilon_0 \in (0,1)$ such that each $f_\xi^* \big |_{\mathbb{B}_\xi(\varepsilon_0) \cap (X'\cup \{\xi\})}$
has a $\sqrt{C^*}$-Lipschitz extension 
\begin{equation} \label{patch} \widehat{f_\xi} : \mathbb{B}_\xi(\varepsilon_0) \longrightarrow Y. \end{equation}
Let $\varepsilon < \varepsilon_0$ be small enough, and $R>1$ large enough, that 
\begin{equation} \label{buffer} \mathrm{(a)} \quad  \frac{C^* +  \varepsilon/\varepsilon_0}{1-  \varepsilon/\varepsilon_0}\leq 1 
\quad\quad \text{and} \quad\quad \mathrm{(b)} \quad
\frac{(C^*+{\Delta}/{R})+2\varepsilon/R}{1-2\varepsilon/R} \leq 1\end{equation}
where $\Delta>0$ still satisfies~\eqref{addloss}.

\begin{lemma} \label{qtree}
Let $\xi, \xi' \in X$ be distance $\geq R$ apart. 
Then $\mathrm{Lip}(G)\leq 1$ for $$G:= f \sqcup \widehat{f_\xi}\big |_{\mathbb{B}_\xi(\varepsilon)} \sqcup \widehat{f_{\xi'}} \big |_{\mathbb{B}_{\xi'}(\varepsilon)}.$$
\end{lemma}
\begin{proof}
Consider $x,x'\in X'\cup \mathbb{B}_\xi(\varepsilon) \cup \mathbb{B}_{\xi'}(\varepsilon)$. We distinguish several cases.

\noindent $\bullet$ (i) If $x,x'\in X'$ then $d(G(x), G(x'))=d(f(x), f(x'))\leq C d(x,x')$ because $f$ is $C$-Lipschitz.

\noindent $\bullet$ (ii) If $x,x'\in \mathbb{B}_\xi(\varepsilon)$ then by construction of $\widehat{f_\xi}$,
\begin{equation} \label{nest}
d(G(x), G(x'))=d(\widehat{f_\xi}(x), \widehat{f_\xi}(x'))\leq \sqrt{C^*} d(x,x'). 
\end{equation}

\noindent $\bullet$ (iii) If $x,x'\in \mathbb{B}_{\xi'}(\varepsilon)$, we do as in (ii), exchanging $\xi$ and $\xi'$. 

\noindent $\bullet$ (iv) If $x\in \mathbb{B}_\xi(\varepsilon)$ and $x'\in X'$, we distinguish two cases:
if $x'\in X'\cap \mathbb{B}_\xi(\varepsilon_0)$, then~\eqref{nest} still applies. If not, then
we compute
$$ \frac{d(G(x'), G(x))}{d(x',x)} 
\leq \frac{d(G(x'), G(\xi)) + d(G(\xi), G(x))}{d(x',\xi) - d(\xi, x)} 
 \leq \frac{C^* d(x',\xi) + d(\xi, x)}{d(x',\xi) - d(\xi, x)} $$
which is $\leq 1$ by~\eqref{buffer}.a, since $d(\xi,x)\leq \varepsilon$ and $d(x',\xi)\geq \varepsilon_0$.

\noindent $\bullet$ (v) If $x\in \mathbb{B}_{\xi'}(\varepsilon)$ and $x'\in X'$, we do as in (iv), exchanging $\xi$ and $\xi'$. 

\noindent $\bullet$ (vi) Up to exchanging $x$ and $x'$, the only remaining case is that $x\in \mathbb{B}_{\xi}(\varepsilon)$ and $x'\in \mathbb{B}_{\xi'}(\varepsilon)$.
It is only here that we will use the assumption $d(\xi, \xi')\geq R$.

We first treat the case $(x,x')=(\xi, \xi')$.
Recall from~\eqref{cxi} the optimal candidates $\eta=G(\xi)$ and $\eta'=G(\xi')$ and optimal constants
$C_\xi, C_{\xi'}<1$ used in the proofs of Theorem~A and Lemma~\ref{1point}. 
By symmetry, we may assume 
\begin{equation} \label{match3}C_{\xi'} \leq C_\xi  \end{equation}
and by definition of $C_{\xi'}$ we have 
\begin{equation} \label{match} {d(\eta',f(z))} \leq C_{\xi'}\, {d(\xi', z)} ~\text{ for all }~z\in X'.\end{equation}
Recall also from~\eqref{xxi} the compact subset $X_\xi \subset X'$, satisfying
\begin{equation} \label{match2} {d(\eta,f(z))} = C_{\xi}\, {d(\xi, z)} ~\text{ for all }~z\in X_\xi. \end{equation}
By Lemma~\ref{1point} we know
\begin{equation} \label{match4}  C_\xi \leq C^* < 1. \end{equation}
Since $\eta$ lies by~\eqref{convhull} in the convex hull of $f(X_\xi)$, we can find $y_1=f(x_1) \in f(X_\xi)$ such that $\widehat{\eta' \eta y_1} \geq \frac{\pi}{2}$. Then, 
\begin{align}
 d(\eta, \eta')  & \leq d(y_1,\eta')-d(y_1, \eta)+\Delta & \text{by \eqref{CAT}.b--\eqref{addloss}} \notag \\ 
 & \leq C_{\xi'} d(x_1,\xi')-C_\xi d(x_1, \xi) + \Delta & \text{by \eqref{match}--\eqref{match2}} \notag \\
 & \leq C_\xi \left ( d(x_1,\xi')-d(x_1, \xi) \right )+ \Delta & \text{by \eqref{match3}} \label{comput} \\
 & \leq C_\xi \, d(\xi', \xi) + \Delta & \text{(triangle inequality)} \notag \\
 & \leq C^* \, d(\xi', \xi) + \Delta & \text{by \eqref{match4}} & . \notag 
\end{align}
 Since by assumption $d(\xi, \xi')\geq R$, it follows that 
\begin{align} 
{d(\eta, \eta')}/{d(\xi, \xi')} & \leq C^*+{\Delta}/{R}  \quad <1 & \text{by \eqref{buffer}.b.}  \label{buffer3} \end{align}
This 
(\footnote{~For $Y$ a tree and $X$ a general metric space, a variant of the computation~\eqref{comput} holds with $\Delta=0$, and a variant of the argument in~\S\ref{prolasch} yields $C_\xi\leq C$.
Taking each $\xi, \xi' \in X \smallsetminus X'$ (independently) to its optimal image $\eta, \eta' \in Y$ therefore produces a global, loss-less extension of $f$: this was proved in~\cite[Th.~B]{las}, as alluded to in the Introduction.}) 
deals with the case $(x,x')=(\xi, \xi')$.

The general case of (vi) is now similar to (iv-v): we can compute
\begin{align*} \frac{d(G(x), G(x'))}{d(x,x')} 
& \leq \frac{d(G(x), \eta) + d(\eta, \eta') + d(\eta', G(x'))}{-d(x,\xi) + d(\xi, \xi') - d(\xi', x')} \\ 
& \leq \frac{(C^* + \Delta/R) d(\xi,\xi') + 2\varepsilon }{d(\xi,\xi') - 2 \varepsilon} & \text{by~\eqref{buffer3}} \\
& = \frac{(C^* + \Delta/R) +{2\varepsilon}/{d(\xi,\xi')}}{1-{2\varepsilon}/{d(\xi,\xi')}} \quad \leq 1 & ~\text{by~\eqref{buffer}}&\mathrm{.b,} \end{align*}
using again $d(\xi, \xi')\geq R$.
Therefore, $\mathrm{Lip}(G)\leq 1$.
\end{proof}

To finish proving Theorem~\ref{main}, consider a maximal $\varepsilon$-sparse subset 
$$\Xi=\big \{\xi_i \big \}_{i\in \mathbb{N}} \subset X.$$ 
This means that the closed balls $\overline{\mathbb{B}_{\xi_i}(\varepsilon)}$ cover $X$ but the $\mathbb{B}_{\xi_i}(\varepsilon/2)$ are pairwise disjoint (i.e.\ the $\xi_i \in \Xi$ are mutually $\geq \varepsilon$ apart).
For example, $\Xi$ can be constructed from a dense sequence $(x_\iota)_{\iota\in \mathbb{N}}$ of $X$ by setting $\xi_1:=x_1$ and letting inductively $\xi_i$ be the first $x_\iota$ lying outside $\mathbb{B}_{\xi_1} (\varepsilon) \cup \dots \cup \mathbb{B}_{\xi_{i-1}} (\varepsilon)$. 

Since $0\geq \kappa_X \geq -1$, the volume of a ball in $X$ is bounded above (resp.\ below) by the volume of a ball of the same radius in hyperbolic space $\mathbb{H}=\mathbb{H}^{\dim(X)}$ (resp.\ in Euclidean space $\mathbb{E}=\mathbb{R}^{\dim(X)}$): indeed, CAT-type inequalities~\eqref{CAT} show that the \emph{Jacobians} of the exponential maps in $\mathbb{H}$, $X$, and $\mathbb{E}$ form, in that order, a weakly decreasing sequence.
Let $N \in \mathbb{N}$ satisfy
\begin{equation} \label{bins}
N \geq \frac{\mathrm{Vol}_{\mathbb{H}}(\mathbb{B}(R+\varepsilon/2))}{\mathrm{Vol}_{\mathbb{E}}(\mathbb{B}(\varepsilon/2))}.
\end{equation}
Each ball $\mathbb{B}_{\xi_i}(R)$ contains at most $N$ points of $\Xi$, because the $\varepsilon/2$-balls centered at those points are disjoint and contained in $\mathbb{B}_{\xi_i}(R+\varepsilon/2)$.
Therefore, we can find a partition of $\Xi$ into ``bins''
$$\Xi=\Xi_1 \sqcup \dots \sqcup \Xi_N$$ 
such that any distinct $\xi, \xi' \in \Xi_j$ satisfy $d(\xi, \xi')\geq R$:
for example, the $\Xi_j$ can be constructed inductively by putting $\xi_1$ in $\Xi_1$, and then dropping in turn each $\xi_i$ into any bin $\Xi_j$ disjoint from $\{\xi_1, \dots, \xi_{i-1} \} \cap \mathbb{B}_{\xi_i}(R)$. 

Recall from~\eqref{patch} the $\sqrt{C^*}$-Lipschitz maps $\widehat{f_{\xi_i}}$ defined in $\varepsilon_0$-neighborhoods of the $\xi_i$.
For each $1\leq j \leq N$, define the map 
$$F_j:= \Big ( \bigsqcup_{\xi\in \Xi_j} \widehat{f_\xi}|_{\mathbb{B}_\xi(\varepsilon)} \sqcup f \Big ) :\quad  \bigcup_{\xi \in \Xi_i} \mathbb{B}_\xi(\varepsilon) \cup X' \longrightarrow Y.$$
By Lemma~\ref{qtree}, since the Lipschitz property can be tested one pair of points at a time, we have in fact $\mathrm{Lip}(F_j) \leq 1$.
By Theorem~A, the $F_j$ admit $1$-Lipschitz extensions $\widehat{F}_j$ to $X$.
Finally we claim that 
\begin{equation} \label{finalC'} F:=\sum_{j=1}^N \frac{1}{N} \widehat{F}_j \: : X \longrightarrow Y
~\text{ satisfies }~
 \mathrm{Lip}(F) \leq 1 - \frac{1-\sqrt{C^*}}{N} =:C' <1. \end{equation}
Indeed, this can be verified in restriction to each ball $\overline{\mathbb{B}_{\xi_i}(\varepsilon)}$ of the covering of~$X$: if $\xi_i$ falls in the bin $\Xi_j$, then on that ball $\widehat{F}_j$ is $\sqrt{C^*}$-Lipschitz by construction while all other $\widehat{F}_{j'}$ are $1$-Lipschitz; we conclude using~\eqref{moyenne}. \qed

\section{Conclusion} \label{seconj}
It seems natural to expect that the lower bound $K$ on curvature, and the upper bound $m$ on dimension, are not necessary in Theorem~\ref{main}.
\begin{conjecture}\label{conj}
For any $C<1$ there exists $C'\in (C,1)$ such that for any Hadamard manifolds $X,Y$ satisfying $\kappa_X\geq -1 \geq \kappa_Y$, every $C$-Lipschitz map from a subset of $X$ to $Y$ has a $C'$-Lipschitz extension to $X$: 
$$\mathcal{L}_{X,Y}(C)\leq C' < 1.$$
\end{conjecture}
\noindent This statement should still hold if both the map and its extension are required to be equivariant under a given pair of actions on $X$ and $Y$: see~\cite{guk}

\smallskip

Loss does occur, i.e.\ $C'>C$ in general, as testified by many examples. 
For instance, since $\ell \mapsto \mathcal{D}_{2\pi/3} (\ell, \ell)$ is strictly convex (see~\eqref{trigo}), a map $f$ that takes just the vertices of, say, a medium-sized equilateral triangle of $\mathbb{H}^2$ to the vertices of a smaller one, cannot be extended 
without loss to the center of the triangle.
In such examples however, the ratio 
${(1-C')}/{(1-C)}$ 
never seems to get very small.
Thus we propose the following strengthening:
\begin{conjecture} \label{conj2}
There exists a universal $\alpha \in (0,1)$ such that $\mathcal{L}_{X,Y}(C) \leq C^\alpha$
\end{conjecture}
Interestingly, this conjecture appears to be open even for $C$ close to $0$.
The article~\cite{lps} shows that $\mathcal{L}_{X,Y}(C)/C$ is bounded above (which for small $C$ is a stronger property), but only under some extra assumptions on the Hadamard manifold~$Y$, such as fixed dimension with pinched curvature.

As $C$ approaches $1$, bounds on the constant $C'$ extracted from our proof of Theorem~\ref{main} are not very stringent. 
Fixing $K \leq -1$ and the dimension, we can estimate~\eqref{arcsinh} for $r=\frac{2\Delta}{1-C}$ to find that $1-C^*$ is on the order of $(1-C)^2$, yielding
$\varepsilon_0 \approx 1-C$, $\varepsilon \approx (1-C)^3$ and crucially $R\approx (1-C)^{-2}$ in~\eqref{buffer}.
In~\eqref{bins} this entails 
$N\approx \mathrm{e}^{-(\Lambda+o(1))/(1-C)^2}$ 
for some $\Lambda>0$, 
hence in~\eqref{finalC'} $$1-C'\approx \mathrm{e}^{-\frac{\Lambda+o(1)}{(1-C)^2}} \quad \text{as } C \rightarrow 1^-,$$
i.e.\ our upper bound $C'$ for $\mathcal{L}_{X,Y}(C)$ is a far cry from Conjecture~\ref{conj2}.


