MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $E$ be homotopy equivalent to a $k$-sphere. Let $q\colon E\to X$ be a map such that given any continuous $f\colon C\to X$ from a compact space $C$, there exists (a non-unique) $\tilde{f}\colon C\to E$ with $q\tilde{f}=f$. Assume also that $X$ is a connected CW complex, but possibly infinite-dimensional.

Can we say anything at all about the homotopy groups of $X$ besides that they must be quotients of the homotopy groups of $\mathbb{S^k}$?

I apologize if this is too easy for the forum, this is not my area.

share|cite|improve this question
up vote 7 down vote accepted

Yes, we can conclude that either $q$ is an equivalence or $X$ is contractible. Since any cycle lives in a compact subset of $X$, $q$ will also induce surjections on homology. It follows that $X$ is a Moore space $M(\mathbb{Z}/n,k)$ for some $n$, and $q$ is homotopy equivalent to the unique map $S^k\to M(\mathbb{Z}/n,k)$ that induces the quotient map $\mathbb{Z}\to\mathbb{Z}/n$ on homology. In particular, $X$ is homotopy equivalent to a finite CW-complex, and lifting such a homotopy equivalence to $E$ we find that $X$ is a retract of $E$ up to homotopy. In particular, this means $\mathbb{Z}\to\mathbb{Z}/n$ must split, so either $n=0$ and $X\simeq S^k$ or $n=1$ and $X$ is contractible.

share|cite|improve this answer
For any $c\in H_n(X)$, there is some compact subset $C\subseteq X$ such that $c$ is in the image of $H_n(C)\to H_n(X)$ (namely, let $C$ be the union of the images of all the simplices appearing in a representative of $c$). The inclusion $C\to X$ lifts to $E$, and hence $c$ also lifts to $H_n(E)$. – Eric Wofsey Aug 3 '14 at 0:25
$C$ is not a simplex but the union of the images of the simplices in $X$. In your $S^1$ example, $C$ would be all of $S^1$, which does not lift to $\mathbb{R}$. – Eric Wofsey Aug 3 '14 at 0:46
I am little confused. Is $C$ a given fixed compact set or for any compact set $C$ and $f$ the lift exists? – Cusp Aug 4 '14 at 6:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.