Hausdorff compact problem

Question

Let $X$ a Tychonoff space and the topological immersion $e: X \to \prod_{s \in S} [0,1]$. For this other question:

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 $.

Now we have the following problem:

Let $L$ a Hausdorff compact and $i: X \to L$ topological immersion with $\overline{i[X]} = L$ and supose that for all Hausdorff compact $K$ and for all continuous function $f: X \to K$, exists only one $g: L \to K$ continuous with $g \circ i = f$. Show that there is an homeomorphism $F: \overline{e(X)} \to L$ with $F \circ e = i$.

My attempt:

By the other question, exists a continuous function $G: \overline{e(X)} \to K$ with $G \circ e = f$. If I could consider $g^{-1}: g(X) \to L$, 'd define $F: e(X) \to L$ for $F = g^{-1} \circ G$. Then as $G \circ e = f = g \circ i$, we'd have $F \circ e = g^{-1} \circ G \circ e = i$. As $e$ and $i$ are topological immersions, $F = i \circ e^{-1}$ is an homeomorphism.

Well, I'm not sure why $g^{-1}$ is well-defined and I could consider it. Can someone give a hint in how proceed this problem??

Thank you

Answer 1

HINT: For convenience let $C=\operatorname{cl}e[X]$. Take $K=C$; the map $e:X\to C$ is continuous, so by hypothesis there is a continuous $f:L\to C$ such that $f\circ i=e$. Similarly, we can take $K=L$ and observe that since $i:X\to L$ is continuous, there is a continuous $g:C\to L$ such that $i=g\circ e$. Note that if we can show that $g$ is a surjective homeomorphism, we’re done, and one way to do this would be to show that $f=g^{-1}$.

Combining results, we see that

$$e=f\circ i=f\circ g\circ e$$

and

$$i=g\circ e=g\circ f\circ i\;.$$

That is, $f\circ g=\operatorname{id}_{e[X]}$, $g\circ f=\operatorname{id}_{i[X]}$. In other words, $f\upharpoonright g\big[e[X]\big]=(g\upharpoonright e[X])^{-1}$, which is at least a step in the right direction. Moreover, $g\circ e=i$, so $g\big[e[X]\big]=i[X]$, and we’ve established that $f\upharpoonright i[X]=(g\upharpoonright e[X])^{-1}$.

• Show that $g[C]$ is closed in $L$ and contains $i[X]$, and conclude that $g[C]=L$, i.e., that $g$ is a surjection.

Now fix $y_0\in C$ and suppose that $(f\circ g)(y_0)=y_1\ne y_0$. Let $U_0$ and $U_1$ be disjoint open nbhds of $y_0$ and $y_1$ such that $\operatorname{cl}_CU_0\cap\operatorname{cl}_CU_1=\varnothing$. (Why is this possible?)

• Show that $\operatorname{cl}_CU_i=\operatorname{cl}_C\big(U_i\cap e[X]\big)$ for $i=0,1$.
• Show that $g(y_0)\in\operatorname{cl}_L\left(g\big[U_0\cap e[X]\big]\right)$ and hence that $y_1\in f\left[\operatorname{cl}_L\left(g\big[U_0\cap e[X]\big]\right)\right]$.
• Conclude that $y_1\in\operatorname{cl}_C\left((f\circ g)\big[U_0\cap e[X]\big]\right)$ and simplify the righthand side to get a contradiction, thereby showing that $f=g^{-1}$.