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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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.


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.