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Intuitively, I can see why this is. I've found a few threads about this, but they only provide, for example, a deformation retraction of $I \times I$ to its diagonal $D = \{ (x,x) \in I \times I \}$, with no further explanation. It's not clear to me exactly how to prove that this yields a deformation retract of $M$ the mobius strip, so that's what I'd like to clarify here.

Let $q$ be the quotient: $I \times I$ $\rightarrow$ $M$. It seems like there are three steps required

  • Show that $\mathbb S^1$ is a subspace of $M$.

  • Find some continuous $r:M\rightarrow \mathbb S^1$, a retraction of $M$.

  • Find a homotopy from $id_M$ to $i \circ r$, where $i$ is the inclusion: $\mathbb S^1 \rightarrow M$

Here's how I think it can be done. Each bullet below corresponds to a bullet above.

  • First, define $f: I \rightarrow D$ by $f(x) = (x,x)$. Then $q\circ f:I\rightarrow M$ makes the same identifications as the quotient $\omega:I\rightarrow\mathbb S^1, \omega = exp(2\pi ix$), so $q(f(I))=q(D)$ is homeomorphic to $\mathbb S^1$. So $\mathbb S^1$ is a subspace of $M$.

  • Second, define $g:I \times I$ $\rightarrow D, (x,y)\mapsto(x,x)$. Then $r=q\circ g:$ $I \times I$ $\rightarrow \mathbb S^1$ is constant on fibers of $q$, so it descends to the quotient to give a continuous map $M\rightarrow \mathbb S^1$. So $\mathbb S^1$ is a retract of $M$.

  • Finally, define the homotopy $H:I^3\rightarrow I^2$ by $(x,y,t)\mapsto (x,y)*(1-t) + (x,x)*t$. This is the homotopy showing that $g$ is a deformation retraction. Define the composite map $q\circ H: I^3\rightarrow M$. Note that $q(H(0,y,t))=q(0, y*(1-t))=q(1, 1-y*(1-t))=q(H(1,1-y,t))$. Since this is constant on fibers of the map $q\times id:I^2 \times I \rightarrow M \times I$, it descends to the quotient, and there is a continuous map $F: M \times I \rightarrow M$ such that $F(m,0) = id_M$ and $F(m,1)$ = $q(D) \approx \mathbb S^1$, so this is the desired homotopy.

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  • $\begingroup$ Do you mean $f(x) = (x,x)$ in your fourth bullet point? Also, in your last bullet point, $H$ is the deformation retraction. $g$ is a retraction. $\endgroup$
    – treble
    Jul 25, 2015 at 3:56
  • $\begingroup$ Regarding f(x), that's right, I'll edit my post. $\endgroup$ Jul 25, 2015 at 4:17
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    $\begingroup$ Be careful about your claim that the map $I^2\times I\to M$ descends to the quotient. You can't take it for granted that $M\times I$, equipped with the product topology, is a quotient of $I^2\times I$ by the map $q^2\times 1_I$. There is some reason, however, why the product $M\times I$ is indeed a quotient of $I^2\times I$, and it has to do with the local compactness of $I$. $\endgroup$ Jul 26, 2015 at 15:11
  • $\begingroup$ $q\times$ $id_I$:$I^2\times I \rightarrow M \times I$ is a continuous and surjective map from a compact space to a Hausdorff space, so it's a quotient by the closed map lemma? $\endgroup$ Jul 26, 2015 at 20:13
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    $\begingroup$ That's right, you can argue like that, provided you have proven that $M$ is Hausdorff. What I'm aiming at, however, is that when $q:X\to Z$ is a quotient map, then $q\times id:X\times Y\to Z\times Y$ is a quotient map whenver $Y$ is a locally compact space. So in particular, $q\times id_I$ is always a quotient map. That means we can define a homotopy on the quotient by defining a homotopy on the base space, no matter what spaces are involved. $\endgroup$ Jul 26, 2015 at 22:45

1 Answer 1

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Yes, this is right. Concisely, the deformation retraction upstairs on $I^2$ descends to a deformation retraction on $M$ via the quotient map $q,$ but one needs to check the details, as you did.

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