# Show that if $X$ is a compact metric space and an ANE, such that $H_n (X) \neq 0$, then $X$ cannot be embedded in $\mathbb{R}^n$.

An absolute neighborhood extensor (ANE) is a space $Y$ such that for every metric space $X$, $A$ - a closed subset of $X$, and a map $f:A \to Y$, there exists an open set $U$ containing $A$ such that $f$ can be extended to a map $U \to Y$. $H_n(X)$ is the $n$th homology group of $X$.

• Usually for homework we ask for what you've tried so far. That way we have a better understanding of what level to phrase the response. Commented Feb 27, 2012 at 14:13
• Mildly related: math.stackexchange.com/questions/106435/… Commented Feb 27, 2012 at 15:12
• To clarify: in the definition of ANE space, there exists such a $U$, right? Commented Feb 27, 2012 at 16:36
• @Steve: The same definition of ANE appears in my linked question. It means that there exists some $U$ to which we can extends the continuous mapping in to $X$. Commented Feb 27, 2012 at 16:37
• Well, not exactly the same definition. In the linked question it is clear it is for some $U$, not necessarily all $U$. Commented Feb 27, 2012 at 16:50

Suppose $i:X \rightarrow \mathbb{R}^n$ is an embedding; since $X$ is compact so is its image $i(X)$. Let $U$ be an open set containing $i(X)$ to which we may extend the map $i^{-1}$. By compactness of $X$ there is a finite union $C$ of simplices containing $X$ and contained in $U$. Since $C$ is a finite union of simplices, it may be triangulated so that it obtains the structure of a finite simplicial complex. Since it is a finite simplicial complex embedded in $\mathbb{R}^n$, we must have $H_n(C)=0$ (in fact, the group of $n$-cycles is already $0$). But now the composition $X \rightarrow C \rightarrow X$ of $i$ and the extension of $i^{-1}$ is equal to the identity of $X$, and hence we get $H_n(C) \neq 0$, contradiction.
• This only works if I interpreted your definition of ANE correctly--- or is it that there exists a neighborhood $U$ for which there is an extension? Commented Feb 27, 2012 at 16:32
• I edited in accordance with the clarification made by the OP. Probably it's true that for any open subset $U$ of $\mathbb{R}^n$, we have $H_n(U)=0$, but I don't quite see how to prove this. Commented Feb 28, 2012 at 13:56
• Hey Steve.. Thanks a lot! May I ask how do you know that such a $C$ exists? And why is it that an embedded finite simplicial complex in ${\mathbb R}^n$ - $C$ must has a trivial n-cycles group? Commented Feb 28, 2012 at 20:00
• For your first question, for each point $p$ of $i(X)$, one may choose a simplex with barycenter $p$ contained in $U$. By compactness of $X$, the interiors of finitely many of these simplices contain $X$, so we may take $C$ to be the union of these finitely many simplices. Commented Feb 28, 2012 at 20:42