# an extension for one of two homotopic maps implies the existence of a homotopic extension for another map.

Consider $A \subset X$ closed and $X$ as a compact metric space and $f,g : A \rightarrow S^{1}$ as two homotopic maps.$f$ has a continuous extension to a map $F:X \rightarrow S^{1}$. prove that $g$ has a continuous extension homotopic with $F$.

Is there an hint?

Thank you very much.

Let $D(Y)$ be the diameter of the set $Y$. Let $t(x)$, for $x \in A$ be the biggest real number such that both $D(F(X\cap B(x,t(x)))) < \frac 13$ and $D(g(A \cap B(x,t(x)))) < \frac 13$. We have $t(x)>0$ for all $x \in X$ thanks to continuity.
Now we will extend the function to a set $T = A \cup (B(x,t(x)/2) \cap X)$. Let $V=(B(x,t(x)/2) \cap X)$. As the image of $V$ by $g$ has a small diameter, it can be identified with $\Re$ , and therefore one can apply Tietze Urysohn extension theorem with the bounded property to get the function $g_2$. As the image of $V$ by $F$ can also be identified to $\Re$, so let's call this function $F_2$ Also, let $h$ be the restriction of the homotopy to $T$. For every $x \in T$ one can lift the path $h(x , \cdot)$. This lifted paths can be joined continuous(easy to see), in order to have a function $h_2$ whose codomain is $\Re$. By combining the functions $F_2$,$h_2$ and $g_2$ over the domain $V \times [0] \cup (V \cap A) \times [0,1] \cup V \times [1]$ (on this order), we can apply Tietze Urysohn extension theorem again. This (when reprojected back to $S_1$) give the extended homotopic function for $T$.
Now we can se the problem changing $g$ by $g_2$ and $A$ by $T$. All the given properties are the same, and as we bounded the extension, the function $t$ is still the same for this new problem. This means the the same $B(x,t(x))$ can be chosen to extend $g_2$. As $X$ is compact, we have a finite cover of this balls, and therefore we can repeat this extension a finite number of times to get the extension over all $X$.