I would like to show that the diagram
$\begin{array}{} A & \stackrel{f}{\longrightarrow} & Y \\ i \downarrow & & \downarrow {\phi_2} \\ X & \stackrel{\phi_1}{\longrightarrow} & X \coprod_f Y \end{array}$
where $i:A \to X$ is an inclusion is a pushout.
Here $A$ is a closed subset of $X$, all maps given are continuous and $X \coprod_f Y$ is the disjoint union $X \coprod Y$ quotient by the equivalence generated by $\{(a,f(a)) \in (X \coprod Y) \times (X \coprod Y): a \in A\}$ (call the equivalence relation $\sim$)
So I start with $\nu: X \coprod Y \to X \coprod_f Y$ (the natural map) and define $\phi_1 = \nu | X$, and $\phi_2 = \nu | Y$.
Then for $a \in A$, $\phi_1(i(a)) = \nu(i(a)) = \nu(f(a)) = \phi_2(f(a))$ and the diagram is commutative.
The other part is to show that this is unique. So let $Q$ be another space such that there exists $\alpha_1:X \to Q$, $\alpha_2:Y \to Q$. We seek a $u: X \coprod_f Y \to Q$
Define the function $\Theta:X\coprod_f Y \to Q$ with $\Theta | X = \alpha_1$ and $\Theta | Y = \alpha_2$.
Then for $a \in A$, $\Theta(i(a)) = \alpha_1(i(a)) = \alpha_2(f(a)) = \Theta(f(a))$.
So this means that $\Theta$ maps elements of the equivalence class $\sim$ from $X \coprod_f Y \to Q$ (maybe I am not saying the last bit clearly, but I think it is clear what I mean!)
Is this all reasonable? I only ask because this whole commutative diagram thing is very new to me...