I am just getting started with category theory and I am trying to prove that
In the category Haus, a continuous map $f:A \rightarrow B$ is an epimorphism iff $f(A)$ is dense in $B$
I found a partial answer here
What are the epimorphisms in the category of Hausdorff spaces?
The proof of "$f(A)$ is dense in $B \implies f$ is an epimorphism" is missing. They only prove the converse I think, which I don't fully understand.
My questions are
1) How do I prove this "$f(A)$ is dense in $B \implies f$ is an epimorphism"?
2) About the proof of the converse which I paste here:
-What is the idea of the proof? How does taking the topological union of two copies of B and the "natural maps" does the trick? I am not sure I understand what a "topological union is", how is it anydifferent from a normal union? By the way, I guess by natural maps they mean this
(taken from https://en.wikipedia.org/wiki/Canonical_map)
-Regarding the highlighted part, could you give more details? Specifically I don't understand how to check that $h \circ f=k\circ f$. Then by the epimorphism property we get $h=k$, how do they conclude that $f[A]$ is dense in $B$? And isn't one more step missing, i.e. showing that $f(A)$ is dense in $B $?
"Morphisms with dense image are epic" is basically another way to say that "if two continuous functions agree on a dense set then they are equal". Actually, let's prove this as a lemma.
Lemma. Let $f, g : X \to Y$ be continuous functions, where $Y$ is Hausdorff. If $\operatorname{Eq}(f,g) := \{x \in X : f(x) = g(x)\}$ is dense in $X$, then $f = g$.
Proof. $\operatorname{Eq}(f,g)$ is the inverse image of $\Delta_Y \subseteq Y \times Y$ under the continuous map $(f,g) : X \to Y \times Y$. Since $Y$ is Hausdorff, $\Delta_Y$ is closed in $Y \times Y$, so by continuity of $(f,g)$ we have that $\operatorname{Eq}(f,g)$ is closed in $X$. If $\operatorname{Eq}(f,g)$ is dense in $X$, we then conclude that $\operatorname{Eq}(f,g) = X$, i.e. $f = g$. $\square$
Now the claim follows easily:
Proposition. Let $h : X \to Y$ be a morphism in $\mathsf{Haus}$ such that $h(X)$ is dense in $Y$. Then $h$ is an epimorphism.
Proof. Let $f,g : Y \to Z$ be morphisms in $\mathsf{Haus}$ such that $f \circ h = g \circ h$. Then $f(y) = g(y)$ for all $y \in h(X)$, i.e. $h(X) \subseteq \operatorname{Eq}(f,g)$. Since $h(X)$ is dense in $Y$, $\operatorname{Eq}(f,g)$ is dense in $Y$. Since $Z$ is Hausdorff, we apply the lemma to conclude that $f = g$. $\square$
Your understanding of the phrase "natural map" is not correct. Here, "disjoint topological union" refers to the coproduct in $\mathsf{Top}$. Given two spaces $A$ and $B$, we may form a space $A \amalg B$, whose underlying set is the disjoint union of $A$ and $B$, and whose open sets are precisely the pairwise disjoint unions of open subsets of $A$ and $B$.
Pictorally, $A \amalg B$ is the topological space which consists of both $A$ and $B$ -- for example, $S^1$ is a circle, and $S^1 \amalg S^1$ is two (disjoint) circles. People will say "natural maps" in this context to refer to the inclusions of $A$ and $B$ into $A \amalg B$.
But the construction in this proof is actually slightly more involved than the coproduct -- it is an amalgamated coproduct. This is where "... corresponding points of the closure of $f[A]$ have been identified" comes in. Here's the whole proof:
Proposition. Let $f : A \to B$ be an epimorphism in $\mathsf{Haus}$. Then $f(A)$ is dense in $B$.
Proof. We begin with an epimorphism $f : A \to B$. Then we form the space $B \amalg B$, which comes with two natural maps $i_1, i_2 : B \to B \amalg B$. Next, we form the quotient $(B \amalg B)/{\sim}$, where $\sim$ is the equivalence relation generated by $i_1(b) \sim i_2(b)$ for all $b \in \overline{f(A)}$. This comes with a natural map (of the kind referenced in your screenshot) $\pi : B \amalg B \to (B \amalg B)/{\sim}$. Now we check that $(B \amalg B)/{\sim}$ is Hausdorff, by noting that $${\sim} = \Delta_{B \amalg B} \cup \operatorname{img}((i_1,i_2) : \overline{f(A)} \to (B \amalg B)^2) \cup \operatorname{img}((i_2,i_1) : \overline{f(A)} \to (B \amalg B)^2)$$ is a closed subset of $(B \amalg B)^2$. Finally, we have that $$(\pi \circ i_1) \circ f = \pi \circ (i_1 \circ f) = \pi \circ (i_2 \circ f) = (\pi \circ i_2) \circ f$$ by construction. Since $f$ is epic by assumption we conclude that $\pi \circ i_1 = \pi \circ i_2$. However, $\pi(i_1(b)) \neq \pi(i_2(b))$ for any $b \in B \setminus \overline{f(A)}$, and so we conclude that $B \setminus \overline{f(A)} = \varnothing$, i.e. $\overline{f(A)} = B$. In other words, $f(A)$ is dense in $B$. $\square$