Urysohn metrization theorem and homeomorphism

Question

Let's check Munkres' “Topology”, page 215, Theorem 34.1 (Urysohn metrization theorem), Step 2. Here are my confusions.

Why is it not enough that $F$ is a bijective and continuous function from $X$ to $Z=F(X)$?

Munkres also proves $F^{-1}$ is continuous from $Z$ to $X$, but I'm confused why it is necessary.

Actually, my question is why homeomorphism of $X$ onto a metric space guarantee $X$ is a metric space (Why $F$ is a bijective and continuous function from $X$ to $Z=F(X)$ cannot guarantee $X$ is a metric space).

---

This a question from Munkres' "Topology". I show you the whole context of the question below.

Answer 1

If you are thinking about pulling back the distance on $Z$ via $F$, that is:

$$ d'(x,y) = d(F(x), F(y)), $$

a bijection is enough. However,in this case it is not guaranteed that the topology induced by $d'$ will be the given topology of $X$.

As an example, think of $X= \mathbb R$ with the discrete topology, $Z=\mathbb R$ with the usual topology and $F=id$. We can put a metric on $X$, but it will not induce the discrete topology.

Notice that $F^{-1}$ continuous is equivalentg to $F$ being open. In this case, given $U\subset X$, $F(U)$ is open in $Z$, and therefore

$$ F(U) = \bigcup_i B(z_i,\epsilon_i) = B(F(x_i),\epsilon_i); $$

hence

$$ U = \bigcup_i B'(x_i,\epsilon_i). $$

(Notice that $F$ must also be continuous, for we want

$$ B'(x,\epsilon) = \{ y\in X \mid d(F(x),F(y)\} = F^{-1}(B(F(x),\epsilon)) $$

to be open.)

Answer 2

Concerning i), if your collection is not infinite then your space has only finitely many open sets and is thereby a finite space with the discrete topology. Why is this the case?

Assume that your space $X$ is infinite. As you have a regular space, in particula a Hausdorff space for any two distinct points $x,y\in X$ you can find neighborhoods $U_x, U_y$ of $x,y$ separating those two points. Now the set $A_x=X\setminus U_y$ is a closed set containing $U_x$, in particular $\overline{U_x}\subset A_x$ and $y\not\in A_x$. So by regularity (using it fully now), you can find open sets $V_x,V_y$ separating $A_x$ and $y$. This construction yields infinitely many distinct open sets $U_x\subset V_x$ such that $\overline{U_x}\subset V_x$ and we can choose them such that they contain a certain point $x$ and not a certain point $y$, so there are infinitely distinct such pairs, so your collection is infinite.

So if your collection is finite your space needs to be finite as well. But any finite $T_1$ space, in particular any finite regular space has the discrete topology. To see this one only notes that $X=\{x_1,...,x_n\}$ and $\{x_i\}$ is closed, because one point sets are closed, and then $\{x_i\}=X\setminus\{x_1,...,x_{i-1},x_{i+1},...x_n\}$ is open because the other set is the finite union of closed sets and thereby closed as well. So any point is open and closed and so the topology is discrete and thereby metrizable with the discrete metric.

Concerning ii) the map has to be a homeomorphism which is only the case if it is inverse is also continuous. See also Urysohn's Metrization Theorem: What is needed to show that $F$ is an embedding?. Well you need an embedding, because there are spaces with a finer topology than a topology coming from a metric that are not metrizable (and they therefor trivially continuously inject into themselves with the coarser metrizable topology), see for example A topology that is finer than a metrizable topology is also metrizable? Of course they will always fail to either have a countable basis or be regular, because else they would be metrizable by the theorem. So one cannot really give a counterexample to the theorem because it is true, bit for that you first have to prove it.