How to construct a homotopy equivalence between a mobius band and a circle?

Question

A mobius band is homotopic equivalent to a circle because the mobius band can deformation retract onto a circle. I am wondering how could we understand this fact from the definition of being homotopic equivalent. How could we construct maps $f:\textrm{mobius band} → S^1$ and $g:S^1 → \textrm{mobius band}$ such that $g∘f ≃ id_{\textrm{mobius band}}$?

Answer 1

Whenever you have a subspace $A\subseteq X$ and a deformation retraction $H:X\times [0,1]\to X$ onto $A$ (such that $H(x,0)=x$ and $H(x,1)\in A$ for all $x\in X$ and $H(a,1)=a$ for all $a\in A$), you get a homotopy equivalence as follows. Let $f:X\to A$ be given by $f(x)=H(x,1)$ and $g:A\to X$ be the inclusion map. Then $f\circ g=Id_A$, and $H$ is a homotopy from $Id_X$ to $g\circ f$.

Answer 2

I try to answer my question. I am pleased if you guys make some corrections.

A mobius band $M$ can be thought of as a quotient of the square $I\times I=[0,1]×[0,1]$ with the ends $\{0\}×[0,1]$ and $\{1\}×[0,1]$ identified by $(0,x)\sim(1,1-x)$. After seeing this intuitive picture, we can continuously shrink the band (over time) to get a circle.

So there is a deformation retract $f_t:M → M, t\in I$, such that $f_0=id_M$ (the identity map on $M$), $f_1(M)=S^1$, and $f_t(s)=s$ for all $s∈S^1$ and $t∈I$.

Now let $g=\iota:S^1\hookrightarrow M$ be the inclusion map and $h:M → S^1$ be a map so that $h = f_1$ Obviously, $h∘g =f_1\circ\iota = id_{S^1}$. On the other hand, since $f_0≃f_1$, we have $f_1≃f_0$. Observe that $f_1:M → M$ is equal to $g∘h:M → M$. So we have $g∘h≃f_0=id_M$. This shows that $M≃S^1$ by definition.