I want to find a surjective map $f:S^1 \to S^1$ which is null-homotopic. Its trivial to see that if a map is surjective then it will definitely be null-homotopic since $S^1 \setminus a$ is homeomorphic to real line but I am looking for a surjective map which is null-homotopic.
I was thinking about the identity map but I cannot extend it to $D^2$ yet, so I am not sure.
Now I am thinking something like
A map from $S^1 \to X$ is nothing but a path in $X$. So choose a path starting at $(0,1)$ then going in the clockwise direction to $(0,1)$ and then coming back from $(0,1)$ back to $(0,1)$ in the anticlockwise direction. Simply put $\gamma$ is a path loop around $(0,1)$ and then I am considering the path $\beta=\gamma * \gamma^{-1}$ where $*$ denotes the join of two paths.
What do you think?
P.S-I think I need a simpler path which can be extended to $D^2 \to D^2$. So I need to choose another simpler path.
Best Answer
The identity map is not nullhomotopic (in $S^1$). If it were, then by definition $S^1$ would be contractible and $\pi_n(S^1)=0$ for all $n$, which is false. Famously, $\pi_1(S^1)=\Bbb Z$.
But, take a generator of $\pi_1(S^1)$: $\gamma:I\to S^1,\,t\mapsto\exp(2\pi it)$ based at $(1,0)$. The loop $\lambda=(\gamma)^{-1}\cdot\gamma$ - in the notation of the fundamental group action - is surjective but it is also nullhomotopic, as you seem to have guessed.
Explicitly: $$\lambda(t)=\begin{cases}\gamma(2t)&0\le t\le\frac{1}{2}\\\gamma(2(1-t))&\frac{1}{2}\le t\le1\end{cases}$$With nullhomotopy given by: $$H:I\times I\to S^1,\,(t,u)\mapsto\begin{cases}\gamma(2t\cdot u)&0\le t\le\frac{1}{2}\\\gamma(2(1-t)\cdot u)&\frac{1}{2}\le t\le 1\end{cases}$$
Note that $H(0,u)=H(1,u)$ for all $u$ and $H(t,0)=(1,0),H(t,1)=\lambda(t)$ for all values of $t$ and $u$. Using $S^1\cong I/\partial I$, we find: $$\lambda:S^1\to S^1$$Which is surjective and: $$H:S^1\times I\to S^1$$A (based) nullhomotopy of $\lambda$ (and $H$ does extend to a map on $D^2$).
Since $D^2\cong S^1\times I/(S^1\times\{0\})$ ($D^2$ is homeomorphic to the cone of $S^1$), by the universal mapping property of a quotient we know $H$ factors through $D^2$. It is a good exercise to explicitly construct the map $D^2\to S^1$ and to explicitly construct the quotient map $S^1\times I\twoheadrightarrow D^2$.