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 $X$.
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 $\mathbb{R}^n$.
6
$\begingroup$
algebraic-topology
-
0Usually 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. – 2012-02-27
-
0Mildly related: http://math.stackexchange.com/questions/106435/prove-that-ane-space-also-has-hep – 2012-02-27
-
0To clarify: in the definition of ANE space, *there exists* such a $U$, right? – 2012-02-27
-
0@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 $X$. – 2012-02-27
-
0Well, not exactly the same definition. In the linked question it is clear it is for some $U$, not necessarily all $U$. – 2012-02-27
-
0In both definitions the intention is that there exists such a U... – 2012-02-27
-
0Well, why not make the intention into reality? Something like "...and a map $f:A \rightarrow Y$, there exists an open set $U$ such that $f$ can be extended to a map $f:U \rightarrow Y$." – 2012-02-28
-
0OK, I hope it is better now.. Steve, any thoughts? – 2012-02-28