I’m looking at some universal properties in the categories of abelian groups, specifically, those of the free products, the direct sums, and abelianization of groups. Most of the statements I’ve seen are along the line of:
Given maps $\phi_i: G_i \to H$,$\dots$, there exists a unique homomorphism $\phi$: [some object built from the $G_i$] $ \to H$ such that [some property is satisfied].
An example is the universal property of the abelianization of groups:
Given a group $G$, an abelian group $H$, and a homomorphism $f: G \to H$, there exists a unique homomorphism $F: G/[G,G] \to H$ such that $f = \pi \circ F$, where $G/[G,G]$ is the quotient of $G$ by its commutator subgroup $[G,G]$, and $\pi$ is the canonical projection.
I’m trying to prove some claims that involves isomorphisms of the groups involved in the above statement. One thing I could think of is to use the first isomorphism theorem. But to do so, I’d need a map $f: G \to H$ that has $[G,G]$ as its kernel (so that $G/[G,G] \cong H$). How do I know that such map exists?
It's not clear to me what you're asking. $G/[G,G]\cong H$ does not always apply, the statement only says there is a homomorphism $G\rightarrow H$ rather than an isomorphism (for example $\mathbb Z\rightarrow0$). All you can say is that $[G,G]\subset\ker f$.
If you want the commutator subgroup to be a kernel: By definition $[G,G]$ is normal, and any normal subgroup forms a quotient, so just take the quotient map $\pi:G\rightarrow G/[G,G]$. But this seems quite unnecessary in your question.
I would not say this is something that uses first isomorphism theorem. It must use the property of abelian group and commutator subgroup. Maybe mimic the proof of first iso theorem, using the value of $f$ vanishes on $[G,G]$, etc. If you really want to use universal properties, I would say universal property of quotients is much better than first iso theorem.