# Homotopy question

Let $X$ be a space and $f:S^1\to X$ be continuous. Show that $f$ is null-homotopic (homotopic to the constant map $e_x:S^1\to${$x$} $\in X$) if and only if there is a continuous map $g:D^2\to X$ with $g|_{S^1}=f$. The hint is to define $g(xt)=F(x,t)$, where $F$ is the homotopy between $f$ and $e_x$, and then use the fact that $(S^1\times [0,1])/$~ is homeomorphic to $D^2$ where the equivalence relation ~ on $S^1\times[0,1]$ is defined as $(x,t)$~$(y,s)$ if and only if $xt=ys$.

I'm really having a hard time keeping track of everything. This is what I've tried:

$f$ is homotopic to $e_x\iff\exists F(x,t):S^1\times[0,1]\to X$ such that $F(x,0)=e_x$ and $F(x,1)=f$. $h:S^1\times[0,1]\to D^2$ by $h(x,t)=xt$ is surjective and continuous (so $h$ is a quotient map onto $D^2$). Now I want to define a map from $g:D^2\to X$ and conclude that the above holds $\iff g:D^2\to X$, the map induced by $F$, is such that $g(x*0)=e_x$ and $g(x*1)=f$.


Then $g$ is continuous by the universal mapping property of quotients. $g|_{S^1}=g(x*1)=f$

I'm sure that I made a mistake somewhere, but could someone please let me know if they see any problems?

## 1 Answer

$F$ induces the map $g$ iff it respects the relation given by $h$, i.e. if $F(x,t)=F(y,s)$ whenever $h(x,t)=xt=ys=h(y,s)$. But these products are equal iff $t=s=0$ (or $(x,t)=(y,s)$), and in this case $F(x,0)=e_x=F(y,0)$. means that $F$ factors as $gh$, and $g$ is continuous since $D^2$ has the final topology for $h$. Also if $z\in\partial D^2$, then $z=h(z,1)$, so $f(z)=F(z,1)=gh(z,1)=g(z)$.

For the other direction (where $g:D^2\to X$ is given), you only need to compose.

In the end, it all boils down to a homeomorphism $$S^1\times I/S\times\{0\}\cong D^2$$ since the maps starting at the left term are exactly the homotopies $F$ from $F|_{S^1\times\{1\}}$ to a constant map at $F(x,0)$.