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While describing Quillen’s $ + $-construction in his book Algebraic K-Theory, V. Srinivas assumes that his topological spaces are equivalent to CW-complexes and are path-connected. As universal covering spaces are featured in the $ + $-construction, and as a topological space has a universal covering space if and only if it is (i) path-connected, (ii) locally path-connected, and (iii) semi-locally simply connected, we are naturally led to the following question:

Is a topological space that is homotopy equivalent to a CW-complex locally path-connected?

I already know that a topological space that is homotopy equivalent to a CW-complex is semi-locally simply connected, but I do not know whether I can replace “semi-locally simply connected” by “locally path-connected” to get an affirmative answer.

Thank you!

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    $\begingroup$ The comb space is a counterexample (it is contractible so trivially of CW type). $\endgroup$
    – Tyrone
    Mar 13 at 18:27
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    $\begingroup$ Let $X$ be connected. Call a covering space of $X$ universal if it is weakly initial in the category of all connected coverings of $X$ (cf. Spanier's Algebraic Topology). If $X$ is locally path-conneced, then any simply connected covering of $X$ is universal (the converse can fail). In fact, a connected, locally-path connected space $X$ has a simply conneced covering space iff $X$ is semilocally simply connected. Nevertheless: any connected space of CW homotopy type has a universal covering space which is simply connected. $\endgroup$
    – Tyrone
    Mar 13 at 18:32
  • $\begingroup$ @Tyrone: Thank you, Tyrone! $\endgroup$ Mar 18 at 21:59

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No. The easiest way to get a counterexample is to consider contractible spaces, which are homotopy equivalent to a single point. For instance, if $X$ is any space at all, then the cone $CX=X\times[0,1]/X\times\{1\}$ is contractible (just contract down to the cone point), but near $X\times\{0\}$ it is locally homeomorphic to $X\times\mathbb{R}$ which is locally path connected iff $X$ is locally path connected.

However, a space that is homotopy equivalent to a space that has a universal cover (defined as a covering space that is simply connected) automatically also has a universal cover. In fact, more strongly, suppose $X$ and $Y$ are path-connected spaces, $f:X\to Y$ is a map that induces an isomorphism on fundamental groups, and $p:\tilde{Y}\to Y$ is a universal cover. Then I claim the pullback $q:\tilde{X}=X\times_Y\tilde{Y}\to X$ of $p$ to $X$ is a universal cover. It is easy to see that since $p$ is a covering map, so is $q$ (if $U\subseteq Y$ is evenly covered by $p$ then $f^{-1}(U)$ is evenly covered by $q$).

To see that $\tilde{X}$ is path-connected, suppose $(a,b),(c,d)\in\tilde{X}$. Since $\tilde{Y}$ is path-connected, there is a path $\gamma$ from $b$ to $d$, and then $p\gamma$ is a path from $p(b)=f(a)$ to $p(d)=f(c)$. Since $f$ induces an isomorphism on fundamental groups, there is a path $\delta$ from $a$ to $c$ such that $f\delta$ is homotopic to $p\gamma$ relative to the endpoints. Then $f\delta$ lifts to a path $\delta'$ in $\tilde{Y}$ which still goes from $b$ to $d$, and $(\delta,\delta')$ defines a path in $\tilde{X}$ from $(a,b)$ to $(c,d)$.

Finally, to see that $\tilde{X}$ is path-connected, it suffices to show that for any non-nullhomotopic loop $\gamma$ in $X$ and any $(a,b)\in\tilde{X}$ such that $a=\gamma(0)$, the lift of $\gamma$ to $\tilde{X}$ starting at $(a,b)$ is not a loop. Since $f$ induces an isomorphism on fundamental groups, $f\gamma$ is non-nullhomotopic, so its lift $\gamma'$ to $\tilde{Y}$ starting at $b$ is not a loop. But now $(\gamma,\gamma')$ is the lift of $\gamma$ to $\tilde{X}$ starting at $(a,b)$, and it is not a loop since $\gamma'$ is not.

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  • $\begingroup$ Thank you, Eric! $\endgroup$ Mar 18 at 21:59

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