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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.

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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.

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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 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 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 at 6:45

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