How does it follow that $P = Q$ in this universal property of coproducts of abelian groups?

Question

How does it follow that $P = Q$ in this universal property of coproducts of abelian groups?

I'm reading Martin Brandenburg's explanation of why:

Given that $P$ is a coproduct of the abelian groups $A_i$, how can we deduce - from the universal property, not from the construction of $P$ - that every element of $P$ is a finite sum of elements coming from the $A_i$?

He says: Now, for the proof, let $Q$ be the subgroup of $P$ generated by the images of the $A_i$. Then the universal property of $P$ induces a canonical homomorphism $f: P \to Q$, and one checks that it is inverse to the inclusion $g: Q \hookrightarrow P$. It follows that $P = Q$.

I simply don't understand how it follows that $P = Q$ or rather why $g \circ f = \mathrm{id}$ on $P$? Can someone please explain?

I get that $f(P) = Q$ where $f$ is surjective. I also get that $f\circ g = \mathrm{id}$ on the domain $Q$ but why should $g \circ f = \mathrm{id}$ on $P$?

Answer 1

Let $\iota_j\colon A_j\to P$ be the canonical injections; let $Q=\langle \iota_j(A_j)\mid j\in I\rangle$ be the subgroup the images generate. Let $\iota'_j\colon A_j \to Q$ be the co-restrictions of the $\iota_j$. Let $g\colon Q\hookrightarrow P$ be the subgroup embedding. Note that $\iota_j = g\circ \iota'_j$.

The morphisms $g\circ \iota'_j\colon A_j\to Q\to P$ induce a unique morphism $\phi\colon P\to P$ such that $\phi\circ\iota_j = g\circ\iota'_j$ for all $j$. Now, it is clear that $\mathrm{id}_P$ satisfies this condition as well (that is, $\mathrm{id}_P\circ\iota_j = g\circ \iota'_j$ for all $j$), so the uniqueness clause of the universal property of $P$ tells you that $\phi=\mathrm{id}_P$.

On the other hand, the morphisms $\iota'_j\colon A_j\to Q$ induce a unique morphism $f\colon P\to Q$ such that $\iota'_j = f\circ\iota_j$ for each $j$. Then $g\circ\iota'_j = g\circ f\circ\iota_j$ holds for each $j$. But that means that $g\circ f$ has the same property as $\phi$, and therefore that $g\circ f = \phi = \mathrm{id}_P$, as desired.

Answer 2

I will proceed with a more abstract answer first.

Consider an arbitrary category $C$. Recall that an arrow $f : A \to B$ is said to be “mono” or “a monomorphism” when for all objects $C$ and arrows $g, h : C \to A$, if $f \circ g = f\circ h$, then $g = h$.

In “concrete categories” (roughly, categories where the arrows between two objects are functions) such as the categories of groups, sets, topological spaces, etc, every injective function is a monomorphism. In most concrete categories, every monomorphism is an injection. This includes all the aforementioned categories. In short, monomorphisms in the category of groups are exactly the injective functions.

An arrow $f : A \to B$ is said to be “split epi” or “a split epimorphism” if there exists some $s : B \to A$ such that $f \circ s = 1_B$. Such an $s$ is known as a “section” of $f$.

Theorem: an arrow is an isomorphism if and only if it is both a monomorphism and a split epimorphism.

Proof: Suppose $f : A \to B$ is an isomorphism; let $g : B \to A$ be its inverse. Then clearly $f \circ g = 1_B$, so $f$ is split epi. Now suppose we have $h, j : C \to A$ and $f \circ h = f \circ j$. Then

$\begin{equation} \begin{split} h &= 1_A \circ h\\ &= (g \circ f) \circ h \\ &= g \circ (f \circ h) \\ &= g \circ (f \circ j) \\ &= (g \circ f) \circ j \\ &= 1_A \circ j \\ &= j \end{split} \end{equation}$

and thus, we see $f$ is mono.

Conversely, let $f$ be mono and split epi. Then let $g : B \to A$ be a section of $f$. Then we have $f \circ g = 1_B$. Then we have

$\begin{equation} \begin{split} f \circ (g \circ f) &= (f \circ g) \circ f \\ &= 1_B \circ f \\ &= f \\ &= f \circ 1_A \end{split} \end{equation}$

Applying the fact that $f$ is mono, we see that $g \circ f = 1_A$. Thus, $f$ is an isomorphism.

Let’s turn back to the problem at hand. We see that the inclusion map $g : Q \to P$ is injective and therefore mono. Moreover, we have produced a map $f : P \to Q$. Making use of the universal property of the coproduct, we will verify that $g \circ f = 1_P$. To do so, we verify that for each inclusion map $a_i : A_i \to P$, we have $g \circ f \circ a_i = a_i$; this fact follows from the definitions of $f$ and $g$.

So we have constructed a mono $g$ and shown that $g$ is split epi; therefore, $g$ is an isomorphism.