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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).

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You're using a definition of "Alexandrov space" very different from the one I'm using (alexandrov <=> all intersections of open sets are open), because with my definition there are no connected alexandrov spaces except the indiscrete spaces. Could you post the definition of alexandrov space you're using? –  Johannes Hahn Apr 15 '10 at 22:48
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It is a geodesic metric space of curvature bounded below in the triangle comparison sense. –  Sergei Ivanov Apr 15 '10 at 22:54
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By the way, these spaces are named after A.D.Alexandrov, and Alexandrov-discrete topologies are named after P.S.Alexandrov. What an unfortunate collision! –  Sergei Ivanov Apr 15 '10 at 23:02
    
The paper by Väisälä, [The invariance of domain under acyclic mappings, Duke Math. J. 33 1966 679--681] seems relevant to your "topological question" It does not give what you want (I think) but it does prove invariance of domain for cohomology manifoldd over $\mathbb Z$. As I do not have electronic access to Duke Math.J., I cannot tell you more about the proof. –  Igor Belegradek Apr 16 '10 at 1:23
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3 Answers

up vote 5 down vote accepted

The following lemma from Grove--Petersen, A radius sphere theorem does the trick.

Lemma 1. Let $X$ be a compact Alexandrov space without boundary. Then $X$ has a fundamental class in Alexander-Spanier cohomology with $\mathbb Z_2$ coefficients; i.e. $\bar H^n(X,\mathbb Z_2 ) = \mathbb Z_2$.


Why: First note that it is true for compact spaces --- in this case the map has $\mathbb Z_2$-degree one. Moreover, in this case the same it true for any continuous map which is injective around one point in the target.

Now let $X$ and $Y$ be $m$-dimensional Alexandrov spaces, $\Omega\subset X$ be an open subset and $f:\Omega\to Y$ be an injective continuous map and $y=f(x)$. One can use $f$ to construct a continuous map between sperical suspensions over spaces of directions $\mathbb S\\,\Sigma_x\to \mathbb S\\,\Sigma_y$ which is injective around one point in the target --- take a small sherical neghborhood $W\ni y$ and collapse everything outside of $W$ to the south pole of $\mathbb S\\,\Sigma_y$. (We can do it sinse small spherical neighborhood of a point $x$ in an Alexandrov space is homeomorphic to cone over space of directions at $x$.)


The same is true for the second question --- link of any pseudomanifold is a pseidomanifold, thus it has $\mathbb Z_2$-fundamental class.

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This is promising, but how do you derive domain invariance from this? I know a proof based on Jordan theorem, and this does not seem to work: the spherical cone over $RP^2$ is not separated by the spherical cone over its equator although both are simply connected. –  Sergei Ivanov Apr 16 '10 at 9:30
    
I'm adding why. –  Anton Petrunin Apr 16 '10 at 19:27
    
How do you map spherical suspensions? For example in the case of $\mathbb R^n$? –  Sergei Ivanov Apr 16 '10 at 19:50
    
@Sergei, is it better now? –  Anton Petrunin Apr 18 '10 at 15:15
    
I can verify the "injective around a point" part in the pseudo-manifold case, by localizing homology near a manifold point and using Domain Invariance for $\mathbb R^n$. In the Alexandrov case, localization of cohomology boils down to the fact that $H^n(X\setminus U)=0$ where $U$ is a small disc around a manifold point. This is probably true, but I don't feel confident in Alexander-Spanier cohomology to be sure. –  Sergei Ivanov Apr 19 '10 at 11:35
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The discussion in the comments is getting too long, let me sum up the proof. This is community wiki; feel free to correct errors and fill in details.

Let $H^*$ denote the Alexander-Spanier cohomology with $\mathbb Z_2$ coefficients. We need the following properties of a compact $n$-dimensional Alexandrov space $X$.

  • $H^n(X)=\mathbb Z_2$. This is Lemma 1 in Grove-Petersen paper, they also rephrase it as "$X$ has a fundamental class"

  • For every manifold point $x\in X$, the map $i^*:H^n(X,X-x)\to H^n(X)$ is an isomorphism. This property is referred to as "having a fundamental class that makes sense".

Hopefully this follows from the fact that $X$ has an open dense connected subset which is a manifold and whose complement has dimension at most $n-3$. Someone who knows something about AS cohomology is needed to clarify this.

We also have $H^{n+1}(X)=0$. This and the second property above and the exact cohomology sequence for the pair $(X,X-x)$ imply that

  • $H^n(X-x)=0$, if $x$ is a manifold point. (Conversely, the second property above follows from this).

We also need the following special case of Perelman's stability theorem:

  • Every point $x\in X$ has a neighborhood homeomorphic to the cone over the space of directions $\Sigma_x$ which is an Alexandrov space of curvature $\ge 1$ and dimension $n-1$, so that the spherical suspension over $\Sigma_x$ is also an $n$-dimensional Alexandrov space. Here $X$ does not need to be compact.

Given all this, the proof works as follows. Let $X,Y$ be $n$-dimensional Alexandrov spaces, $U\subset X$ an open set, $f:U\to Y$ an injective map, $x\in U$ and $y=f(x)$. We are to prove that $f(U)$ contains a neighborhood of $y$. We can make $U$ so small that its closure is compact and $f$ extends to this closure.

Let $U'\subset\subset U$ be a small cone neighborhood of $x$ and $C=U-U'$. Then $U/C$ is homeomorphic to the spherical suspension over $\Sigma_x$ so that it satisfies the above nice properties. Let $D\subset Y$ be a complement of a cone neighborhood of $y$, where the neighborhood is so small that $f(C)$ does not touch it. Now we have a new map $f:U/C\to Y/D$ which is injective on $f^{-1}(Y-D)$. It suffices to show that the new $f$ is onto, and we only need to prove that its image covers all manifold points.

There exists a manifold point $x'\in U-C$ such that $y':=f(x')\in Y-D$ and $y'$ is also a manifold point. By degree theory (which btw depends on domain invariance in $\mathbb R^n$), the map $f^*:H^n(U/C,U/C-x')\to H^n(Y/D,Y/D-y')$ is an isomorphism. By the second property above it follows that $f^*: H^n(U/C)\to H^n(Y/D)$ is an isomorphism.

Now suppose that there exists $z\in Y-D$ which is not in the image of $f$. Then $f$ can be filtered through $Y/D-z$, but $H^n(Y/D-z)=0$, so $f^*: H^n(U/C)\to H^n(Y/D)$ is zero, a contradiction.

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Sergei, It seems you are convinced that it is true, but you really want to have a clean proof --- that probably means that you have an application. Can I ask what is it? –  Anton Petrunin Apr 21 '10 at 17:21
    
I was asked this question by Yuri Burago who was asked by another guy whose name I forgot. I just want to handle him all details I can. –  Sergei Ivanov Apr 21 '10 at 19:04
    
For Alexandrov spaces (or more generally for Hausdorff paracompact locally contractible spaces) Alexander-Spanier and singular cohomology coincide, as shown in Chapter 6, Section 9, Theorem 1 of Spanier's "Algebraic topology". However this does not simplify matters; how does one prove $H^n(X-x)=0$ for singular cohomology? This is worrisome. –  Igor Belegradek Apr 22 '10 at 1:23
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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 $H^n(X-U)=0$.

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