Let $D^2$ be closed unit disk and $S^1$ be its boundary. Prove or disprove:there is an extension of a continuous map $f:S^1 \rightarrow S^1$ to a continuous map $F:D^2 \rightarrow S^1$ iff $f$ is null-homotopic.
My thinking is that the statement is incorrect.If $f$ is null-homotopic,then $X$ is contractible,which means $X$ is simply connected.Thus $\pi_1(S^1)$ is trivial,which is a contradiction since we know i should be $\mathbb{Z}$.
On
Your thinking isn't right. The function $f:S^1\to S^1$ given by $f(x, y) = (1,0)$ is nullhomotopic, and it's quite easy to extend it to a continuous function $F:D^1\to S^1$. Just because $f$ represents the $0$ element in $\pi_1(S^1)$, that doesn't mean that $\pi_1(S^1)$ is zero.
Actually, since $f$ is nullhomotopic, think about what that means: There is a continuous function $H:S^1\times [0,1]\to S^1$ such that $H(x, 0) = f(x)$, and $x\mapsto H(x, 1)$ is a constant function. So $f$ can be extended to a function $H$ defined on a cylinder so that one of the edges of the cylinder is mapped to a single point in $S^1$. We are very close to a solution now. Can you transform this $H$ into the $F$ we are asked about?
In general, this argument shows that a nullhomotopy of a function $S^1\to X$ is equivalent to a continuous extension to $D^1\to X$, which is the main reason why the fundamental group and the first homotopy group are so colsely related.
If $f$ is null-homotopic then there exists $H: S^1\times [0,1]\to S^1$ such that $H(x,0)=f$ and $H(x,1)=const$
but you can define a continuos map $G:D^2\to S^1\times [0,1]$ such that $G(x)=H(x,0) $ for each $x\in S^1\subseteq D^2$ so $F:=H\circ G$
If $f$ can be extended then there exist $F: D^2\to S^1$ such that $F\circ i=f$ where $i: S^1\to D^2$ is the inclusion map. Now you can define a continuos map $K: S^1\times [0,1]\to D^2$ such that $K(x,0)=f(x)$ for each $x\in S^1$ while $K(x,1)=F(0)$. In this case you have that $H:= F\circ K$ is an homoty from $f$ to the constant function $F(0)$ so $f$ is null-homotopic.