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If $X$ and $Y$ are two homotopy equivalent space, and $X$ is semilocally simply connected (equivalently, $X$ has a universal cover) must $Y$ be semilocally simply connected? How would one prove/find a counterexample for this statement? I tried constructing a counterexample, but I really only know of two spaces which are not semilocally simply connected (the shrinking wedge of circles and $\mathbb R^2$ with the "$K$-topology") and these don't seem to be homotopy equivalent to anything nicer.

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I was never much one for finding pathalogical examples, but it seems that it suffices to show that $Y$ does not have a covering space. Thus the question becomes: if $g:Y \to X$ is a homotopy equivalence of spaces and $X$ has a universal cover, then does $Y$? What if we just take the pull-back of the diagram? –  Juan S Sep 25 '12 at 10:07
    
I don't think this works. The homotopy equivalence maps can be very degenerate, so they need not induce a universal cover on $Y$. –  user15464 Sep 25 '12 at 13:58

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Suppose $Y$ is semilocally simply connected and $f:X \to Y$ is a homotopy equivalence. We want to show $X$ is semilocally simply connected, i.e. that for all $x \in X$ there is an open neighborhood $U$ such that the map $\pi_1(U,x) \to \pi_1(X,x)$ is trivial.

Since $Y$ is semilocally simply connected, there is an open neighborhood $V$ of $f(x)$ such that the map $\pi_1(V,f(x)) \to \pi_1(Y,f(x))$ is zero. Let $U = f^{-1}(V)$, a neighborhood of $x$.

Then the two maps $$ \pi_1(U,x) \to \pi_1(V,f(x)) \stackrel{triv}{\to} \pi_1(Y,f(x)) $$ (which is trivial) and $$ \pi_1(U,x) \to \pi_1(X,x) \stackrel{\sim}{\to} \pi_1(Y,f(x)) $$ coincide, showing that the map $\pi_1(U,x) \to \pi_1(X,x)$ is trivial as desired.

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