Endometry Lemma Let $X$ be a proper metric space with a cocompact isometry group. Let $f : X\to X$ be an isometric map. Then how to see $f$ is bijective?
Since $f$ is isometric, it's easy to see $f$ is injective. But how to see $f$ is surjective?
 A: Here is an argument in the general case. 
Let $G$ be the isometry group of $X$; by assumption closed bounded subsets of $X$ are compact (properness assumption) and the isometry group $G$ satisfies: $M:=\sup_{x,z\in X}d(x,Gz)<\infty$ (cocompactness assumption).
For $r>0,m\ge 0$ and $x\in X$, define $u_{r,m}(x)\ge 0$ as the largest number of points at pairwise distance $\le r$ in the closed $(r+m)$-ball around $x$. By properness, $u_{r,m}(x)<\infty$, and $u_{r,m}$ is bounded on bounded subsets. Also, $u_{r,m}(gx)=u_{r,m}(x)$ for every $g\in G$, which implies that $N_{r,m}:=\sup_{x\in X} u_{r,m}(x)<\infty$. Also, $u_{r,m}$ is lower semicontinuous (again using properness); hence $Z_{r,m}=u_{r,m}^{-1}(\{N_{r,m}\})$ is non-empty, closed, $G$-invariant. 
By contradiction, suppose that $X$ has a non-surjective isometric self-embedding $f$. Choose $y\notin f(X)$ and $R=d(y,f(X))>0$ (note that $f(X)$ is complete, hence closed). 
By properness, there exists $x'\in X$ with $d(y,f(x'))=R$. There exists $z\in Z_{R,M}$ with $d(z,x')\le M$. Hence, writing $N=N_{R,M}$, there exists $N$ points $x_1,\dots,x_N$ pairwise at distance $\ge R$, with $d(x_i,z)\le R+M$. Thus $f(x_1),\dots,f(x_N)$ are pairwise at distance $\ge R$, withe $d(f(x_i),f(z))\le R+M$. In addition $d(y,f(x_i))\ge R$ for every $i$, and $d(y,f(z))\le d(y,f(x'))+d(f(x'),f(z))\le R+M$. We deduce that $u_{R,M}(f(z))\ge N+1$, a contradiction.
A: I can only prove the case when $X$ is a homogeneous space, i.e. the isometry group acts transitively on $X$.
In that case, by composing $f$ with a bijective isometry, one can assume that $f$ has a fixed point $x$, therefore maps any closed ball $B(x, d)$ to itself.
Since $X$ is proper, the ball $B(x, d)$ is compact and hence $f$ is surjective on $B(x, d)$ (c.f. this question).
The radius $d$ being arbitrary, we see that $f$ is surjective on $X$.

I'm not intelligent enough to cook up a solution to the general case, but hope this answer helps!
P.S.: It seems that I find the source: at page 31 and there's a hint there...
