Suppose $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. I'd like to prove that if $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$.

I've had the following thoughts:

Let $f : X \to Y$ and $g: Y \to X$ be such that $gf \simeq \mathrm{id}_X$ and $fg \simeq \mathrm{id}_Y$, and let $p : \tilde{X} \to X$, $q : \tilde{Y} \to Y$ be the covering projections. Since $Y$ is locally path connected, so is $\tilde{Y}$. There is a Lemma that says:

Suppose $\pi : B \to A$ is a covering projection, and $F: C \to A$ is a continuous map where $C$ is simply connected and locally path connected. Suppose given base points $a_0, b_0, c_0$ of $A,B,C$ with $\pi(b_0) = a_0 = F(c_0)$. Then there is a unique continuous $\tilde{F} : C \to B$ with $\pi \tilde{F} = F$ and $\tilde{F}(c_0) = b_0$.

I'm going to use this Lemma with $C = \tilde{Y}$ and $F = fq$. So pick points $x_0 \in X$, $\tilde{x_0} \in \tilde{X}$, $\tilde{y_0} \in Y$ such that $p(\tilde{x_0}) = x_0 = fq(\tilde{y_0})$. Then $fq$ has a unique lifting to a map $\tilde{f}: \tilde{Y} \to \tilde{X}$ such that $p \tilde{f} = fq$ and $\tilde{f}(\tilde{y_0}) = \tilde{x_0}$. Similarly, $gp$ has a unique lifting to a map $\tilde{g} : \tilde{X} \to \tilde{Y}$ such that $\tilde{g} = gp$ and $\tilde{g}(\tilde{x_0}) = \tilde{y_0}$.

I'd like it to be the case that $\tilde{f}$ and $\tilde{g}$ are homotopy equivalences. I don't know if this is true and, if it is, I don't know how to show it.

I can see that $p \tilde{f} \tilde{g} \simeq p$ and $q \tilde{g} \tilde{f} \simeq q$ which is quite close to what I want, but I don't know how to proceed.

I'm also concerned that I haven't use the "path connected" criterion anywhere. Hints would be great!


  • 1
    $\begingroup$ See also here and here. $\endgroup$ – t.b. May 22 '12 at 19:10
  • $\begingroup$ @t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p \tilde{f} \tilde{g} \simeq p$ to $\tilde{f} \tilde{g} \simeq \mathrm{id}_{\tilde{X}}$? $\endgroup$ – Matt May 22 '12 at 19:15
  • $\begingroup$ By construction $\tilde{f}\tilde{g}$ is a deck transformation of a simply connected space, hence it must be homotopic to the identity ($\pi_1 \tilde{X} = 1$). $\endgroup$ – t.b. May 22 '12 at 19:26
  • $\begingroup$ @t.b. Actually, sorry. Why is $\tilde{f} \tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$. $\endgroup$ – Matt May 22 '12 at 19:43
  • $\begingroup$ You probably mean $pf = p$, no? That's what Clive verifies right before that. $\endgroup$ – t.b. May 22 '12 at 19:56

Assume that $\phi:X \longrightarrow Y$ is a homeomorphism. The map $ \tilde{X} \longrightarrow X \longrightarrow Y$ can be (uniquely) lifted to a map $\Phi:\tilde{X} \longrightarrow \tilde{Y}$. See the image.

enter image description here

Now do a similar map with $\tilde{Y} \longrightarrow Y \longrightarrow X$ (the vertical map.)

Notice that $\Psi \circ \Phi : \tilde{X} \longrightarrow \tilde{Y}$ is a lift of $p: \tilde{X} \longrightarrow X$. By uniqueness of lifts, this has got to be the identity map.

  • 1
    $\begingroup$ $X$ and $Y$ aren't homeomorphic, they're homotopy equivalent. $\endgroup$ – Najib Idrissi Mar 1 '18 at 8:58
  • $\begingroup$ Oh! You're right. But I think the same argument works by lifting the maps (pointwise) for all 0<t<1 of the homotopy. I have not verified it, though. $\endgroup$ – Behnam Esmayli Mar 2 '18 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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