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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!

Thanks

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    $\begingroup$ 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
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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.

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    $\begingroup$ $X$ 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

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