Question about null-homotopic function

Question

How to prove that every continuous function $f:M\to X$ from the Möbius strip $M$ into a simply connected space $X$ is null-homotopic? Thanks in advance.

Answer 1

Since $M \simeq S^1$, take WLOG $f:S^1 \to X$. By definition a loop in $X$ is a path $g:I \to X$ such that $g(0)=g(1)$. Notice $S^1=I/\sim, 0\sim1$. So any map $f:S^1 \to X$ is just a loop and induces a loop in $\pi_1(X)$.

$\pi_1(X)=0$ so the loop has to be homotopic to a constant (trivial) loop and hence by definition null-homotopic.

Answer 2

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} $$

The Möbius strip $M$ is the quotient space of $I^2$ obtained by the identification $(0,y)\sim(1,1-y)$. \begin{array}{ccc} I^2\times I &\ra d &I^2\\ \da{q\times Id} & &\da q\\ M\times I &\ra{\tilde d} &M \end{array} The square $I^2$ deformation retracts onto $I\times\left\{\frac12\right\}$ via $d((x,y),t)=tr(x,y)+(1-t)(x,y)=\left(x,\ \frac t2+(1-t)y\right)$ where $r(x,y)=\left(x,\frac12\right)$ is the retraction. Since $d((0,y),t)\sim d((x,1-y),t)$ (i.e. it respects the identification) and $q\times Id$ is a quotient map, it induces a deformation retraction $\tilde d$ of $M$ onto $\left(I\times\left\{\frac12\right\}\right)/\sim\ \approx S^1$. Composing $\tilde d$ with $f$ we obtain a homotopy between $f$ and $f\circ r$, the latter being the image of a circle in $X$. This circle can be contracted in $X$ since it is simply connected. Concatenating these homotopies gives a homotopy between $f$ and a constant map.