This question comes from section 44 problem 4 of Munkres.

Let $X$ be a Hausdorff space. Let $I=[0,1]$. Show that if there is a continuous surjective map $f : I \rightarrow X$, then $X$ is compact, connected, weakly locally connected, and metrizable. [Hint: Show $f$ is a perfect map.]

I have already shown everything except that it is metrizable. I know that since $X$ is compact it is also paracompact, so I was going to use the Smirnov metrization theorem which states that if $X$ is paracompact and locally metrizable that $X$ is metrizable.

I am struggling with how to set up the neighborhood $U$ of $x\in X$, such that $U$ is metrizable. Any hints would be great in this direction.

Alternatively I thought about proving that $X$ was second-countable but I was also struggling with that. So any hints this direction would also be welcome.

## Best Answer

Sorry, my first answer is wrong!!! Please allow me to correct the answer.

Situation.$X$ is Hausdorff, $(Y,d)$ is a compact metric space, and $f:Y \to X$ is a continuous surjection.Then $X$ is compact.

Definition.Let $n\geq 1$ be an integer and $x,x'\in X$ a point. Write $$ \tilde{F}_n := \{ (y_1,y_2)\in Y\times Y | d(f^{-1}(f(y_1)),y_2) \geq 1/n\}. $$ Then $\tilde{F}_n\subset Y\times Y$ is a closed symmetric subset. Write $\Delta \subset X\times X$ for the diagonal.Claim 1.Proof.If $f(y_1) = f(y_2)$, then $y_2\in f^{-1}(f(y_1))$. Hence $d(f^{-1}(f(y_1)),y_2) = 0$. This implies the first equality. The second equality follows from the compactness of $f^{-1}(f(y))$ (for any $y\in Y$). The third equality follows from the second equality. $\square$Definition.For any $\tilde{A}\subset X\times X$, we write $\tilde{A}^{\mathrm{op}} := \{(a,b)|(b,a)\in \tilde{A}\}$, and $\tilde{A}^{\mathrm{sym}} := \tilde{A}\cap \tilde{A}^{\mathrm{op}}$. Write $\tilde{W}_n:= (X\times X \setminus (f\times f)(F_n))^{\mathrm{sym}}$. Then $\tilde{W}_n$ is open and symmetric. Write $$ \mathcal{U}_n := \{U \ | \ U\text{ is open}, U\times U \subset \tilde{W}_n\} $$ for the open covering of $X$. Write \begin{align*} U_n(x) &:= \{ x'\in X | \forall y'\in f^{-1}(x'), \exists y\in f^{-1}(x), d(y,y') < 1/n\}, \\ W_n(x) &:= \{x'\in X | (x,x')\in \tilde{W}_n\}. \end{align*}Claim 2.For any $x\in X$ and any open subset $x\in U\subset X$, there exists $n\geq 1$ such that $U_n(x)\subset U$.Proof.Take an integer $n \gg 1$ such that $d(f^{-1}(x),f^{-1}(X\setminus U)) > 1/n$. Then $U_n(x) \subset U$. $\square$Claim 3.$W_n(x) \subset U_n(x)$.Proof.Assume that $(x,x')\in W_n$. Then $(f\times f)^{-1}(x,x') \cap F_n = \emptyset$. Hence for any $y'\in f^{-1}(x')$, $d(f^{-1}(x),y') < 1/n$, i.e., $\exists y\in f^{-1}(x), d(y,y') < 1/n$. This implites that $x'\in U_n(x)$. $\square$Claim 4.If $U\times U \subset W_n$, and $x\in U$, then $U\subset W_n(x)$.Proof.If $x'\in U$, then $(x,x')\in U\times U \subset W_n$. Thus $x'\in W_n(x)$. $\square$Definition.Since $X$ is compact (hence in particular, paracompact), we can take a finite star (open) refinement $\mathcal{V}_n$ of the open covering $\{U\cap V | U\in \mathcal{U}_n, V\in \mathcal{V}_{n-1}\}$, inductively. Write $$\mathcal{V}_n(x) := \bigcup \{V\in \mathcal{V}_n | x\in V\} \subset X.$$ Then it follows immediately that $\mathcal{V}_n(x) \subset \mathcal{V}_{n-1}(x)$. Finally, we write $\mathcal{V} := \bigcup_{n\geq 1}\mathcal{V}_n$. Then $\mathcal{V}$ is a countable set of open subsets of $X$.Claim 5.$\mathcal{V}$ is a basis of $X$. In particular, $X$ is second countable.Proof.Let $x\in X$ be a point and $x\in U\subset X$ an open neighborhood of $x$. By Claim 2 and Claim 3, there exists an integer $n\geq 1$ such that $W_n(x) \subset U$. Since $\mathcal{V}_n$ is a star refinement of $\mathcal{U}_n$, $\mathcal{V}_n(x) \times \mathcal{V}_n(x) \subset W_n$. Hence, by Claim 4, $\mathcal{V}_n(x) \subset W_n(x) \subset U$. This implies that $\exists V\in \mathcal{V}_n$ such that $x\in V \subset U$. $\square$Conclusion.$X$ is metrizable.Proof.This follows directly from Uryshon metrization theorem. $\square$