Showing that there is a surjective map from $\Bbb Z \ast \Bbb Z$ to $C_2 \ast C_3$ just using universal property of coproduct

Question

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.

Answer 1

First, let $T:\mathbf{C}\to\mathbf{D}$ be a left-adjoint covariant functor from a category $\mathbf{C}$ to a category $\mathbf{D}$, that is, there exists a covariant functor $S:\mathbf{D}\to\mathbf{C}$ such that $$\text{hom}_\mathbf{D}\big(T(x),y\big)\cong\text{hom}_\mathbf{C}\big(x,S(y)\big)\text{ for all }x\in\mathbf{C}\text{ and }y\in\mathbf{D}\,.$$ We claim that $T$ preserves epimorphisms. Suppose that $f:x\to x'$ is an epimorphism of objects $x,x'\in\mathbf{C}$. Then, $T(f):T(x)\to T(x')$ is an epimorphism if and only if $$\text{hom}_\mathbf{D}\big(T(x'),y\big)\overset{F}{\longrightarrow }\text{hom}_\mathbf{D}\big(T(x),y\big)$$ is an injective function, where $F(\phi):=\phi\circ T(f)$ for all $\phi\in \text{hom}_\mathbf{D}\big(T(x'),y\big)$. 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 $F'=\psi\circ f$ for all $\psi\in \text{hom}_\mathbf{C}\big(x',S(y)\big)$. 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 $\prod\limits_{i\in I}\,\mathbf{D}$. That is, the objects of $\mathbf{C}$ are families $(G_i)_{i\in I}$ of objects in $\mathbf{D}$ and morphisms from an object $(G_i)_{i\in I}$ of $\mathbf{C}$ to an object $(H_i)_{i\in I}$ of $\mathbf{C}$ are the families $(\phi_i)_{i\in I}$ of morphisms $\phi_i:G_i\to H_i$ in $\mathbf{D}$. Note that $(\phi_i)_{i\in I}$ is an epimorphism in $\mathbf{C}$ if and only if each $\phi_i$ is an epimorphism.

Take $T:\mathbf{C}\to\mathbf{D}$ to be the $I$-coproduct functor: $$T\Big(\left(G_i\right)_{i\in I}\Big):=\coprod_{i\in I}\,G_i\text{ for all }G_i\in \mathbf{D}\text{ where }i\in I\,.$$ Then, $T$ is a left-adjoint covariant functor, with the right adjoint being $S:\mathbf{D}\to\mathbf{C}$ defined by $$S(G):=(G)_{i\in I}\text{ for each }G\in \textbf{D}\,.$$ 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.

Answer 2

Do you mean something like this? Let $\phi_1:G_1\to H_1$ and $\phi_2:G_2\to H_2$ 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 $$\phi(g_1^1g_2^1g_1^2g_2^2\cdots g_1^kg_2^k):=\phi_1(g_1^1)\,\phi_2(g_2^1)\,\phi_1(g_1^2)\,\phi_2(g_2^2)\,\cdots \,\phi_1(g_1^k)\,\phi_2(g_2^k)$$ for all $g_1^1,g_1^2,\ldots,g_1^k\in G_1$ and $g_2^1,g_2^2,\ldots,g_2^k\in G_2$. 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 $G_1\to (G_1*G_2)$ and $G_2\to (G_1*G_2)$, respectively. Similarly, $i_1:H_1\to (H_1*H_2)$ and $i_2:H_2\to(H_1*H_2)$ are the canonical injections. We note that there exist maps $\psi_1:=i_1\circ \phi_1:G_1\to (H_1*H_2)$ and $\psi_2:=i_2\circ\phi_2:G_2\to (H_1*H_2)$. By universality of coproducts, there exists a unique map $\phi:(G_1*G_2)\to (H_1*H_2)$ such that $\phi\circ\iota_1=\psi_1$ and $\phi\circ\iota_2=\psi_2$. Check that $\phi$ is the map given above. The rest (regarding injectivity or surjectivity of $\phi$) is trivial.