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I am solving Allufi chapter $0$ exercise $3.7$. There is a easy way to solve this if we know how the coproduct of $Bbb Z ast Bbb Z$ and $C_2 ast C_3$. I was wondering if there is an abstract approach to solving this without knowing information about the groups.

First, let $T:mathbf{C}tomathbf{D}$ be a left-adjoint covariant functor from a category $mathbf{C}$ to a category $mathbf{D}$, that is, there exists a covariant functor $S:mathbf{D}tomathbf{C}$ such that $$text{hom}_mathbf{D}big(T(x),ybig)congtext{hom}_mathbf{C}big(x,S(y)big)text{ for all }xinmathbf{C}text{ and }yinmathbf{D},.$$
We claim that $T$ preserves epimorphisms. Suppose that $f:xto x'$ is an epimorphism of objects $x,x'inmathbf{C}$. Then, $T(f):T(x)to T(x')$ is an epimorphism if and only if
$$text{hom}_mathbf{D}big(T(x'),ybig)overset{F}{longrightarrow }text{hom}_mathbf{D}big(T(x),ybig)$$
is an injective function, where $F(phi):=phicirc T(f)$ for all $phiin text{hom}_mathbf{D}big(T(x'),ybig)$. By left-adjointness of $T$, $F$ induces a map
$$text{hom}_mathbf{C}big(x',S(y)big)overset{F'}{longrightarrow}text{hom}_mathbf{C}(x,S(y)big),,$$
where $F'=psicirc f$ for all $psiin text{hom}_mathbf{C}big(x',S(y)big)$. As $f$ is an epimorphism, $F'$ is an injective function, and so is $F$. Therefore, $T(f)$ is also an epimorphism.

Fix an index set $I$. Let $mathbf{D}$ be the category of groups (or any category that admits coproducts on the index set $I$), and $mathbf{C}$ the product category $prodlimits_{iin I},mathbf{D}$. That is, the objects of $mathbf{C}$ are families $(G_i)_{iin I}$ of objects in $mathbf{D}$ and morphisms from an object $(G_i)_{iin I}$ of $mathbf{C}$ to an object $(H_i)_{iin I}$ of $mathbf{C}$ are the families $(phi_i)_{iin I}$ of morphisms $phi_i:G_ito H_i$ in $mathbf{D}$. Note that $(phi_i)_{iin I}$ is an epimorphism in $mathbf{C}$ if and only if each $phi_i$ is an epimorphism.

Take $T:mathbf{C}tomathbf{D}$ to be the $I$-coproduct functor:
$$TBig(left(G_iright)_{iin I}Big):=coprod_{iin I},G_itext{ for all }G_iin mathbf{D}text{ where }iin I,.$$
Then, $T$ is a left-adjoint covariant functor, with the right adjoint being $S:mathbf{D}tomathbf{C}$ defined by
$$S(G):=(G)_{iin I}text{ for each }Gin textbf{D},.$$
Thus, $T$ preserves epimorphisms.

On the other hand, it is an unrelated happy coincidence that $T$ also preserves monomorphisms in the case $mathbf{D}$ is the category of groups. This is not true in the general setting. For reference, see here.

Do you mean something like this? Let $phi_1:G_1to H_1$ and $phi_2:G_2to H_2$ be homomorphisms of groups $G_1$, $G_2$, $H_1$, and $H_2$. Then, $phi_1$ and $phi_2$ induces a group homomorphism
$$phi:=phi_1*phi_2:(G_1*G_2)to (H_1*H_2)$$
defined by
$$phi(g_1^1g_2^1g_1^2g_2^2cdots g_1^kg_2^k):=phi_1(g_1^1),phi_2(g_2^1),phi_1(g_1^2),phi_2(g_2^2),cdots ,phi_1(g_1^k),phi_2(g_2^k)$$
for all $g_1^1,g_1^2,ldots,g_1^kin G_1$ and $g_2^1,g_2^2,ldots,g_2^kin G_2$. The map $phi$ is injective if and only if both $phi_1$ and $phi_2$ are injective. The map $phi$ is surjective if and only if both $phi_1$ and $phi_2$ are surjective.

Proof. Let $iota_1$ and $iota_2$ denote the canonical injections $G_1to (G_1*G_2)$ and $G_2to (G_1*G_2)$, respectively. Similarly, $i_1:H_1to (H_1*H_2)$ and $i_2:H_2to(H_1*H_2)$ are the canonical injections. We note that there exist maps $psi_1:=i_1circ phi_1:G_1to (H_1*H_2)$ and $psi_2:=i_2circphi_2:G_2to (H_1*H_2)$. By universality of coproducts, there exists a unique map $phi:(G_1*G_2)to (H_1*H_2)$ such that $phicirciota_1=psi_1$ and $phicirciota_2=psi_2$. Check that $phi$ is the map given above. The rest (regarding injectivity or surjectivity of $phi$) is trivial.