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

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!

• The comb space is a counterexample (it is contractible so trivially of CW type). Mar 13 at 18:27
• 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. Mar 13 at 18:32
• @Tyrone: Thank you, Tyrone! Mar 18 at 21:59

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.