Proof:

Let us begin with Gr and a given pair of morphisms $j_1:C\longrightarrow A_1$ and $j_2:C\longrightarrow A_2$. Let $G$ be the coproduct of the groups $A_1$ and $A_2$. We regard $A_1$ and $A_2$ as subgroups of $G$. Let $N$ be the normal subgroup of $G$ generated by

$$\{j_1(c)j_2(c)^{-1}/ c\in C\}$$

and $\eta:G\longrightarrow G/N$ be the quotient map. We claim that $G/N$ qualifies as the push-out with the associated homomorphisms

$$f_1 = \eta\circ i_1,\quad f_2 = \eta\circ i_2$$

where $i_1$ and $i_2$ are the inclusions of $A_1, A_2$ in $G$. Since $\eta(j_1(c)) = \eta(j_2(c))$, we see that $f_1\circ j_1 = f_2\circ j_2$. To check the universal property, let $g_1:A_1\longrightarrow H$ and $g_2:A_2\longrightarrow H$ be a pair of morphisms such that

$$g_1\circ j_1 = g_2\circ j_2 \eqno(23.10)$$

Aside from (23.10), by definition of coproduct, there exists a unique homomorphism $\psi:G\longrightarrow H$ such that $\psi\circ i_1 = g_1$ and $\psi\circ i_2 = g_2$ from which follows easily that the kernel of $\psi$ contains $N$. Let $\overline{\psi} : G/N\longrightarrow H$ be the unique map such that $\overline{\psi}\circ \eta = \psi$. Then

$$\overline{\psi}\circ\eta\circ i_1 = \psi\circ i_1 = g_1,\quad \overline{\psi}\circ\eta\circ i_2 = g_2.$$

which means $\overline{\psi}\circ f_1 = g_1$ and $\overline{\psi}\circ f_2 = g_2$. That completes the job of verifying that $G/N$ is indeed the push-out. Note that we have only used the definition of coproducts and the most basic property of quotients. As a result the proof goes through verbatim for the other two situations as we shall see. Leaving aside the case of abelian groups we pass on to the category Top.

Well, changing notations to suit the need, let $h_1:Z\longrightarrow X$ and $h_2:Z\longrightarrow Y$ be a pair of continuous functions and $X\sqcup Y$ be the disjoint union of $X$ and $Y$, and $i_1:X\longrightarrow X\sqcup Y$, $i_2:Y\longrightarrow X\sqcup Y$ be the canonical inclusions. For each $z \in Z$ we identify the points $(i_1\circ h_1)(z)$ and $(i_2\circ h_2)(z)$ in $X\sqcup Y$ and $W$ be the quotient space with the projection map

$$\eta:X\sqcup Y\longrightarrow W = (X\sqcup Y)/\sim$$

We claim that $W$ qualifies to be the push-out with associated morphisms $f_1 = \eta\circ i_1:X\longrightarrow W$ and $f_2 = \eta\circ i_2:Y\longrightarrow W$. To check the first condition observe that since $(i_1\circ h_1)(z)$ and $(i_2\circ h_2)(z)$ are identified, $\eta(i_1(h_1(z))) = \eta(i_2(h_2(z)))$ which means $f_1\circ h_1 = f_2\circ h_2$. Turning now to the universal property let $g_1:X\longrightarrow T$ and $g_2:Y\longrightarrow T$ be two continuous maps such that

$$g_1\circ h_1 = g_2\circ h_2. \eqno(23.11)$$

Aside from (23.11), since $X\sqcup Y$ is the coproduct in Top, there is a unique continuous map $\psi:X\sqcup Y\longrightarrow T$ such that $\psi\circ i_1 = g_1$ and $\psi\circ i_2 = g_2$. Now (23.11) implies that $\psi$ respects the identification and so there is a unique $\overline{\psi}:(X\sqcup Y)/\sim\;\longrightarrow T$ such that $\overline{\psi}\circ \eta = \psi$. By the universal property of the quotient, $\overline{\psi}$ is continuous and

$$\overline{\psi}\circ f_1 = \overline\psi\circ\eta\circ i_1 = \psi\circ i_1 = g_1,$$

and likewise $\overline{\psi}\circ f_2 = g_2$. That suffices for a proof.