Let $X$ be a topological space and $f:S^{1}\rightarrow X$ continuous. Show that f is null homotopic if and only if there exist a continuous function $g:D^2\rightarrow X$ with $g$ restricted to $S^1$ equal to $f$.
Let $F:S^1\times I \rightarrow X$ be a homotopy bewteen a constant map $c$ and $f$, that is
$F(x,0)=c(x)$
$F(x,1)=f(x)$
I defined $g(z)=g(xt)=F(x,t)$. Here, $D^2=\{z\in \mathbb{C}:||z||\le 1\}$ and therefore, every such $z$ can be written as $z=xt$ for $x\in S^1$ and $t \in I=[0,1]$. $g$ is continuous because is a composition of $f$ and the identity map betweeen $D^2$ and $S^1$ also g verifies the restriction condition because of the homotopy.
Any hints on the other direction? thanks
Define $r : S^1 \times I \to D^2, r(z,t) = tz$. This is a continuous surjection between compact Hausdorff spaces, hence a quotient map. We have $r(z,0) = 0$ and $r(z,1) = z$ for all $z \in S^1$.
If $f$ has a continuous extension $g : D^2 \to X$, then $F = g r : S^1 \times I \to X$ is the desired homotopy. In fact, $F(z,0) = g(r(z,0)) = g(0)$ and $F(z,1) = g(r(z,1)) = g(z) = f(z)$.
Conversely, if there exists a homotopy $F : c \simeq f$ as in your question, then we get a function $g : D^2 \to X$ by defining $g(0) = c$ and $g(z) = F(z/\lvert z \rvert, \lvert z \rvert)$ for $z \ne 0$. We have $gr(z,0) = g(0) = c = F(z,0)$ and $gr(z,t) = g(zt) = F(tz/\lvert tz \rvert, \lvert tz \rvert) = F(z,t)$ for $t \ne 0$. This shows $gr = F$. The universal property of quotient maps implies that $g$ is continuous.