TITLE: Is there Domain Invariance for Alexandrov spaces?
QUESTION [16 upvotes]: A colleague asked me this question recently. Every injective continuous map between manifolds of the same (finite) dimension is open - this is Brouwer's Domain Invariance Theorem. Is the same true for complete boundaryless Alexandrov spaces (of curvature bounded below)?
Alexandrov spaces are manifolds almost everywhere, and their singularities have special structure. In dimensions 1 and 2 there are no topological singularities (all Alexandrov spaces are manifolds). Higher dimensional singularities have a sort of inductive description: every point $x$ in an $n$-dimensional Alexandrov space has a neighborhood homeomorphic to the cone over an $(n-1)$-dimensional connected(!) Alexandrov space $\Sigma_x$ which carries a metric of curvature $\ge 1$. The last property implies that $\Sigma_x$ is compact and its fundamental group is finite.
For example, in dimension 3 the only type of singularity is the cone over $RP^2$. In dimension 4 there are cones over lens spaces, $\mathbb R\times Cone(RP^2)$ and maybe other beasts.
There is a similar purely topological question: is Domain Invariance Theorem true for "almost manifolds"? An "almost manifold" is a pseudo-manifold obtained from $n$-simlices by gluing their $(n-1)$-dimensional faces in pairs - that is, there are exactly two $n$-simplices adjacent to each $(n-1)$-simplex, and there are no extra identifications between lower-dimensional faces. I'm not sure that all Alexandrov spaces admit triangulations, but if they do, they are "almost manifolds" (of a special kind).

REPLY [6 votes]: Here I will clarify the cohomological issues in Sergei's answer above. For applications to Alexandrov spaces scroll to the end of the post.
I will use Alexander-Spanier cohomology with compact support and $\mathbb Z_2$ coefficients, and the main reference will be Massey's book "Homology and cohomology theory, an approach based on Alexander-Spanier cochains"; I own a Russian translation with insightful comments by Sklyarenko. As usual, using $\mathbb Z_2$ coefficients allows to ignore orientability issues.
Incidentally, as is explained in Massey's book (or in Spanier's "Algebraic topology" text) for locally contractible, locally compact Hausdorff spaces (e.g. for finite-dimensional Alexandrov spaces) the  Alexander-Spanier cohomology coincide with singular and Cech cohomology. 
Lemma. Let $X$ be a locally compact Hausdorff space that 
contains a closed subset $S$ such that $X-S$ is a connected topological $n$-manifold. 
If $U$ is an open subset of $X-S$, then the homomorphism $H_c^n(X, X-U)\to H_c^n(X,S)$ induced by inclusion is an isomorphism.
Proof. By Theorem 1.4 in Chapter 1 of Massey's book, if $A$ is a closed subset of $X$, then there is an isomorphism $H^n_c(X,A)\cong H^n_c(X-A)$.
Theorem 3.21 in Chapter 3 of Massey's book says that if $U$ is an open subset of a manifold $M$,
then the map $H^n_c(U)\to H^n_c(M)$, which associates to a cocycle with compact support in $U$ the same cocycle with support in $M$, is an isomorphism. 
Look at the inclusion $(X, S)\to (X, X-U)$.
Using the above isomorphism, we can identify the induced map $H^n_c(X, X-U)\to H^n_c(X,S)$ with $H^n_c(U)\to H^n_c(X-S)$, which is an isomorphism, as $U$ is open in the manifold $X-S$. QED
Below we denote by $H^n$ the Alexander-Spanier cohomology with arbitrary support;
they coincide with singular cohomology for nice spaces, such as locally contractible, locally compact Hausdorff spaces. Of course, for compact $X$ cohomology with compact support coincide with the the usual cohomology, so we get:
Corollary. If in the assumptions of the Lemma $X$ is compact, then the map
$H^n(X, X-U)\to H^n(X, S)$ induced by inclusion is an isomorphism. QED
Finally, as in Grove-Peterson's paper from Anton's answer, if $X$ is a compact $n$-dimensional Alexandrov space without boundary, and $S$ is the set of non-manifold points, then $S$ has codimension $2$, so long exact sequence of the pair $(X,S)$ shows that $H^n(X,S)\to H^n(X)$ is an isomorphism by inclusion, and we get the isomorphism $H^n(X, X-U)\to H^n(X)$
which implies $H^n(X-U)=0$ by the exact sequence of the pair $(X, X-U)$ because all cohomology in dimension $>n$ vanish. In summary:

If $X$ is a compact $n$-dimensional Alexandrov space without boundary, and $S$ is the set of non-manifold points, then $H^n(X-U)=0$ for any open subset $U$ of $X-S$.

Now if $x$ is a point of $X-S$, then $X-x$
deformation retracts to some $X-U$, so $H^n(X-x)=0$ which is what's needed for Sergei's answer. 
Remark. In fact, the above assertion that $H^n(X-U)=0$ holds for any open $U$ in $X$, i.e. we need not assume $U\subset X-S$. Indeed, if $V:=U-S$, then the isomorphism $H^n(X, X-V)\to H^n(X)$ factors through
the the homomorphism $H^n(X, X-U)\to H^n(X)$, so the latter is onto, but its cokernel is $H^n(X-U)$, hence $H^n(X-U)=0$.