How to show that a diagram is a pushout in the category $\text{TOP}$?

Question

Show that the diagram: $$\require{AMScd} \begin{CD} A @>f>> Y \\ @ViVV @VVV \\ X @>>> X \amalg_f Y \end{CD}$$ where $i : A \hookrightarrow X$ is an inclusion is a pushout in $\mathbf{Top}$.

I can prove that $g_1 : Y \rightarrow X \coprod_fY$, and $g_2 : X \rightarrow X\coprod_f Y$ is a solution of the graph, but I'm going in circles trying to show that any additional solution $(D, h_1, h_2)$ must have a unique morphism $\phi : X\coprod_f Y \rightarrow D$ that makes the diagram commute.

$X \coprod_f Y = (X \coprod Y) / \sim$, where $\sim$ is the equivalence relation generated by $\{(a,f(a)) \in (X \coprod Y) \times (X \coprod Y): a \in A, \text{ where $A$ is a closed subset of $X$} \}$

Anyone have any ideas?

Answer 1

Note that by definition, the maps $X,Y \rightrightarrows X \sqcup_f Y$ are induced by the inclusions $X,Y \rightrightarrows X \sqcup Y$. Suppose you have another solution $(Z,u,v)$, and a (continuous) map $\phi:X \sqcup_f Y \to Z$ making the diagram commute. Composing with the quotient map we get a map $\bar\phi : X \sqcup Y \to Z$ or, equivalently, two maps $X,Y \rightrightarrows Z$. Commutativity of the diagram then dictates that $\bar\phi (x) = u(x)$ and $\bar\phi(y) = v(y)$ for all $x \in X, y \in Y$. This means that the only way to define $\phi$ is to let $\phi([x]) = u(x)$ and $\phi([y]) = v(y)$. Moreover, since $u \circ i = v \circ f$, we have $$\phi([f(a)]) = v(a) = u(a) = \phi([a])$$ for any $a \in A$, so $\phi$ is a well-defined continuous map.

Note that it doesn't matter whether $A \subset X$ is closed or not. (But in case it is closed, the parallel map $Y \to X \sqcup_f Y$ is also a closed embedding.)