Show that for all compact $K$ and for all continuous function $f:X \to K$, there is $g: \overline{e(X)} \to K$ continuous with $g \circ e = f $.

Question

Let $X$ a Tychonoff space, $S = ${$f:X \to [0,1] : f$ continuous} and consider the topology immersion $e: X \to \prod_{s \in S} [0,1]$ where $e(x) = (f(x))_{f \in S}, \quad \forall x \in X$.

Show that for all compact $K$ and for all continuous function $f:X \to K$, there is $g: \overline{e(X)} \to K$ continuous with $g \circ e = f $.

Some observations: By Tychonoff Theorem, $[0,1]^S$ is compact and Hausdorff, so $\overline {e(X)}$ is compact and Hausdorff, then it's Normal and Regular.

Can someone give a hint for this question?

Thank you

Answer 1

Embed $K$ into a product $[0,1]^I$ for some $I$ ($I$ can have the size as a base for $K$), say via $i: K \rightarrow [0,1]^I$, where $i: K \rightarrow i[K]$ is a homeomorphism.

Then consider the maps $p_i:= \pi_i \circ i \circ f$ for all $i$. The $p_i$ are among the functions in $S$. Then $g_i = \pi_{p_i}: \overline{e[X]} \rightarrow [0,1]$ restricted to $\overline{e[X]}$ obeys $g_i \circ e = p_i$. Then $\prod_i g_i$ maps $\overline{e[X]}$ to $[0,1]^I$, and in fact to $i[K]$. Now pull back via $i$, as this is a homeomorphism.