Previous: [[Consequences of adjoint functors preserving limits]] To give another example to chew on and make some of the examples from before clearer, I'll discuss the functor $\operatorname{ab}: \mathbf{Grp} \to \mathbf{Ab}$ from the category of groups to the category of abelian groups. To each group $G$ it assigns $G^{ab}$, the "freest" abelian group built out of $G$. We'll discuss the construction, and see how it behaves with respect to some of the things we have discussed in the past. It ties together many concepts we've talked about, and is semi-important later. If you're not particularly interested in the category theory, you can probably skim this one. If we have a group $G$ that's not necessarily abelian, what is the most "natural" way of making it abelian? One idea is to "force it to be abelian" by forcibly commuting the elements. In [[Quotients of free groups]] we discussed the idea of "adding relations to a group," and how it is the same as quotienting by the maximum normal subgroup generated by those relations. Well, let's express $G$ via generators and relations, and then for every $g,h \in G$ we manually add the relation $gh = hg \iff ghg^{-1}h^{-1} = e$. The subgroup $\langle ghg^{-1}h^{-1} \forall g, h \in G\rangle$ is called the *commutator* subgroup of $G$, and is denoted $[G, G]$. One can [prove](https://math.stackexchange.com/questions/141888/how-to-show-that-the-commutator-subgroup-is-a-normal-subgroup) that it is in fact normal, so then adding in all the "abelianization relations" is just quotienting by this subgroup. Hence, we define the *abelianization* of $G$ via $G^{ab} = G/[G, G]$ Abelianization has the following universal property: If $\phi: G \to A$ is any map from $G$ into an *abelian* group, then in fact $\phi$ factors through $G^{ab}$ i.e. there is a unique map $\phi^{ab}: G^{ab} \to A$ such that $\phi = \phi^{ab} \circ \pi^{ab}_G$ where $\pi^{ab}_G$ is the projection onto the quotient. The diagram here is <iframe class="quiver-embed" src="https://q.uiver.app/#q=WzAsMyxbMCwwLCJHIixbMCwwLDEwMCwxXV0sWzAsMSwiR157YWJ9IixbMCwwLDEwMCwxXV0sWzEsMCwiQSIsWzAsMCwxMDAsMV1dLFswLDIsIlxccGhpIiwwLHsiY29sb3VyIjpbMCwwLDEwMF19LFswLDAsMTAwLDFdXSxbMCwxLCJcXHBpXnthYn1fRyIsMix7ImNvbG91ciI6WzAsMCwxMDBdfSxbMCwwLDEwMCwxXV0sWzEsMiwiXFxwaGlee2FifSIsMix7ImNvbG91ciI6WzAsMCwxMDBdLCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19LFswLDAsMTAwLDFdXV0=&embed" width="304" height="304" style="border-radius: 8px; border: none;"></iframe> To prove this, notice that since the image is abelian, $\phi(a)\phi(b) = \phi(b)\phi(a)$ so $[G, G] \le \ker(\phi)$. Then, define $\phi^{ab}([g])$ to be $\phi(g)$; this is well defined because if $[g] = [h]$ then $gh^{-1} \in [G, G]$ so $\phi(g)\phi(h)^{-1} = e$. Notice that we have given a mapping from the category $\mathbf{Grp}$ to the category $\mathbf{Ab}$ of abelian groups, such mappings qualify to be functors if they satisfy certain additional hypotheses. Let's check that $-^{ab}$ is in fact a functor. If $G$ and $H$ are groups and $\phi: G\to H$ then postcomposing with the projection $\phi_H: H \to H/[H, H]$ gives us a map $\phi': G \to H^{ab}$. Since $H^{ab}$ is abelian there is a unique $\phi^{ab}$ from $G^{ab} \to H^{ab}$. It even makes this diagram commute^[come back to this comment after reading the whole post. You may notice that this is also the data of a natural transformation from the identity functor to the abelianization. This is precisely the beginning of the theory of *monads* on categories, which require some more conditions to be fulfilled. Since inclusion and abelianization form an adjunction, there does exist a monad associated to that adjunction that is at play here]: <iframe class="quiver-embed" src="https://q.uiver.app/#q=WzAsNCxbMCwwLCJHIixbMCwwLDEwMCwxXV0sWzEsMCwiSCIsWzAsMCwxMDAsMV1dLFswLDEsIkdee2FifSIsWzAsMCwxMDAsMV1dLFsxLDEsIkhee2FifSIsWzAsMCwxMDAsMV1dLFsyLDMsIlxccGhpXnthYn0iLDIseyJjb2xvdXIiOlswLDAsMTAwXX0sWzAsMCwxMDAsMV1dLFswLDIsIlxccGlfRyIsMix7ImNvbG91ciI6WzAsMCwxMDBdfSxbMCwwLDEwMCwxXV0sWzEsMywiXFxwaV9IIiwwLHsiY29sb3VyIjpbMCwwLDEwMF19LFswLDAsMTAwLDFdXSxbMCwxLCJcXHBoaSIsMCx7ImNvbG91ciI6WzAsMCwxMDBdfSxbMCwwLDEwMCwxXV1d&embed" width="304" height="304" style="border-radius: 8px; border: none;"></iframe> If we let $G = H$ and $\phi = \operatorname{id}$ we'll see that the only possibility for $\phi^{ab}$ is the identity via this diagram. To check composition, we use the existence of this diagram and construct <iframe class="quiver-embed" src="https://q.uiver.app/#q=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&embed" width="432" height="304" style="border-radius: 8px; border: none;"></iframe> Thus abelianization really is a functor, and all the abstract nonsense about functors applies. To go any further, we would have to talk about the notion of a "subcategory." A subcategory $\mathcal S$ of $\mathcal C$ is just a category formed by taking some of the objects of $\mathcal C$ and some of the arrows, such that the only arrows we took have domain and codomain in $\mathcal S$. Intuitively, from $\mathcal C$ take some objects, and take some of the arrows between the objects we already took to not have any "floating" arrows. For example, we have the subcategory of finite sets inside the category of sets. A subcategory is *full* if we took along all the morphisms we could have. That is, for any arrow in $\mathcal C$ whose domain and codomain is both in $\mathcal S$, the arrow is also in $S$. Again, the subcategory of finite sets and all maps between them is an example. A nonexample is the subcategory of sets where the only arrows are isomorphisms. The reason we talked about this is that $\mathbf{Ab}$ is a full subcategory of $\mathbf{Grp}$. From this subcategory we have an inclusion functor $i$ to $\mathbf{Grp}$ that sends an abelian group to the underlying group. This is very much like a forgetful functor - it forgets that our group is abelian. Then we can see that our universal property implies exactly that $\operatorname{Hom}_{\mathbf{Ab}}(G^{ab}, H) \cong \operatorname{Hom}_{\mathbf{Grp}}(G, i(H))$ If we think of the inclusion functor as a "forgetful" functor that forgets structure, then the abelianization functor is a "free" functor that adds the most general structure that it can to any group that may not have that structure. A subcategory for which such a functor exists, i.e., a subcategory in which the inclusion functor has a left-adjoint, is called a *reflective subcategory* and the adjoint is called a *reflection.* It should be thought of as something that adds the "freest" extra structure required to map into the subcategory. An example in which this idea is clearer, which I'll let you work out, is that of the subcategory of fields in the category of integral domains and *injective* maps between them. Here, the inclusion functor is clear, and the reflection is given by the *field-of-fractions* functor which "completes" a given integral domain by adding in the most minimal structure required to turn it into a field; the classic example is $\operatorname{Frac}(\mathbb Z) = \mathbb Q$. See [Wikipedia](https://en.wikipedia.org/wiki/Field_of_fractions). Since the abelianization functor is a left-adjoint, it preserves colimits^[Confusing? Here's a mnemonic. RAPL: Right Adjoints Preserve Limits. Then the other thing must be true for the other one] and in particular the coproduct. Remember that the coproduct in the category of abelian groups is *not* the free product (that would not even be abelian!) but the direct sum. Thus, $(G\ast H)^{ab} \cong G^{ab}\oplus H^{ab}$ One can also see this relation via using the universal properties. Any map $\phi: G\ast H \to A$ for some abelian group $A$ gives rise to a unique $\phi^{ab}: (G \ast H)^{ab} \to A$ via the property of abelianization, but also gives rise to two unique $\phi_G: G \to A$ and $\phi_H: H \to A$ via the universal property of the free product. Then we can get $\phi_G^{ab}: G^{ab}\to A$ and $\phi_H^{ab}: H^{ab}\to A$ where we use the universal property of the direct sum to get a map from $G^{ab} \oplus H^{ab} \to A$. Then we can draw the obvious diagram^[just put all the objects I talked about in a blank space and connect them with the arrows I mentioned, checking that it commutes using the universal properties I mentioned. This is a nice exercise, and were it not very late, and this post somewhat optional, I would have made it myself] and see that everything commutes giving us a correspondence between $\phi^{ab}$ and $\phi_G^{ab} \oplus \phi_H^{ab}$. Alternatively, see [this mathstackexchange post](https://math.stackexchange.com/questions/1736138/abelianization-of-free-product-is-the-direct-sum-of-abelianizations).^[if there was a nice way to get the align environment working in obsidian I would have written this out too] I must say the "proof via abstract nonsense" invoking the fact that RAPL and LAPC is nicer, at the cost of the fact that we didn't actually do the semi-involved math required to actually *prove* that statement^[someday, when I'm less focused on these slightly less category heavy posts] Next: [[Subgroups of free groups, Part I]]