How to construct a homotopy equivalence between a mobius band and a circle? 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 $g∘f ≃ id_{\textrm{mobius band}}$?
 A: 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 $g\circ f$.
A: I try to answer my question. I am pleased if you guys make some corrections.
A mobius band can be think of a square $[0,1]×[0,1]$ identified in both ends ${0}×[0,1]$ and ${1}×[0,1]$ by $(0,x)$~$(1,1-x)$. After seeing this picture, we can intuitively shrink the band from time to time to get a circle. So there is a deformation retract $f_t:M → M, t∈I$, $M$ for mobius band such that $f_0$ is 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 $g:S^1 → M$ be an inclusion map. Let $h:M → S^1$ be a map such that $h = f_1$ Obviously, $h∘g = id_{S^1}$. On the other hand, since $f_0≃f_1$ , $f_1≃f_0$. Observe that $f_1:M → M$ is equal to $g∘h:M → M$. So we have $g∘h≃f_0$ which is the identity map on $M$. This should show that $M≃S^1$ by definition. 
