Regular monomorphism in Top category

Question

I want to prove that regular monomorphism in Top are exactly the embedding.

We say that a injective continuous function $f: X \to Y$ between topological spaces is an embedding iff $\tau_X = \{f^{-1}(V): V \in \tau_{\,Y}\}$.

We say the a monomorphism $f: X \to Y$ in a category is regular is it arises as the equalizer of to morphism $g,h: Y \to Z$.

Now, given an embedding $f: X \to Y$, I consider $Z = \{0,1\}$ with the indescrete topology and the continuous functions $g= 1$ (the constant function) and $h(y)= 1$ iff $y \in f(X)$ (the characteristic function of the set $f(X)$). Clearly, $gf=hf$.

How to prove the universal property of equalizer? If for some $f':X' \to Y$ I have $gf'=hf'$, then $f'(x')\in f(X)$, for all $x'\in X'$. Thus $f'(x')=f(x)$ for some unique (by injectivity of $f$) $x \in X$. So that $f^{-1}\circ f'$ make the triangle commutes. Is it a continuous mapping? I now $f'$ is continuous by hp. What about $f^{-1}: f(X) \to X$

Conversely, it suffices to prove that the equalizer of two parallel continuous functions $g,h: Y \to Z$ is just the equalizer set $\{y \in Y:g(y)=h(y)\}$ endowed with the usual subspace topology. Also in this case, I can I prove the universal property?

Thanks

Answer 1

1. Let's write $f_1:X\to f(X)$ for $f$, i.e. $f_1(x):=f(x)$, and thus $f$ is the composite of $f_1$ and the inclusion $f(X)\hookrightarrow Y$.
   Now, we have that $f$ is an embedding if and only if $f_1$ is a homeomorphism. (Note that $f^{-1}(V)=f_1^{-1}(V\cap f(X))$ for any $V\subseteq Y$.)
   This proves that $f_1^{-1}\circ f'$ is continuous.
   To conclude the universal property, uniqueness should also be proven, but now given that $f$ is injective - and hence mono -, it comes immediately.

2. Let $X$ denote the equalizer subspace $X:=\{y\in Y:g(y)=h(y)\}$, and let $f:A\to Y$ equalize $g$ and $h$, i.e. $g\circ f=h\circ f$.
   Then for all $a\in A$, we have $g(f(a))=h(f(a))$, so that $f(a)\in X$. Since we consider the subspace topology on $X$, we can easily check that the same $f$ defines a continuous $A\to X$, and composing it with the inclusion $X\hookrightarrow Y$ gives $f$.
   Uniqueness for the universal property follows again because the inclusion is mono.