$\newcommand{Ab}{\operatorname{Ab}} \newcommand{Id}{\operatorname{Id}}$I'm self-studying Introduction to Topological Manifolds by John M. Lee, which includes quite a few exercises like this:

9-4(b) Let $S_1$ and $S_2$ be disjoint sets, and let $R_i$ be a subset of the free group $F(S_i)$ for $i=1,2$. Prove that $\langle S_1 \cup S_2 \mid R_1 \cup R_2 \rangle$ is a presentation of the free product group $\langle S_1 \mid R_1 \rangle * \langle S_2 \mid R_2 \rangle$.

10-17. For any groups $G_1$ and $G_2$, show that $\Ab(G_1*G_2) \cong \Ab(G_1) \oplus \Ab(G_2)$.

10-19. For any set $S$, show that the abelianization of the free group $F(S)$ is isomorphic to the free abelian group $\mathbb{Z}S$.

($*$ is free product, $\Ab$ is abelianization.) Here is my tedious proof of 10-17:

For $i=1,2$, let \begin{align*} \alpha_{i}:G_{i} & \to\Ab(G_{i}),\\ \alpha:G_{1}*G_{2} & \to\Ab(G_{1}*G_{2}),\\ j_{i}:\Ab(G_{i}) & \to\Ab(G_{1})\oplus\Ab(G_{2}),\\ k_{i}:G_{i} & \to G_{1}*G_{2} \end{align*} be the canonical maps. There exists a homomorphism $\ell:G_{1}*G_{2}\to\Ab(G_{1})\oplus\Ab(G_{2})$ satisfying $\ell\circ k_{i}=j_{i}\circ\alpha_{i}$, and there exists a homomorphism $\varphi:\Ab(G_{1}*G_{2})\to\Ab(G_{1})\oplus\Ab(G_{2})$ satisfying $\varphi\circ\alpha=\ell$. Also, there exist homomorphisms $m_{i}:\Ab(G_{i})\to\Ab(G_{1}*G_{2})$ satisfying $m_{i}\circ\alpha_{i}=\alpha\circ k_{i}$, so there exists a homomorphism $\psi:\Ab(G_{1})\oplus\Ab(G_{2})\to\Ab(G_{1}*G_{2})$ satisfying $\psi\circ j_{i}=m_{i}$. Now $$ \varphi\circ\psi\circ j_{i}\circ\alpha_{i}=\varphi\circ m_{i}\circ\alpha_{i}=\varphi\circ\alpha\circ k_{i}=\ell\circ k_{i}=j_{i}\circ\alpha_{i}, $$ so $\varphi\circ\psi=\Id_{\Ab(G_{1})\oplus\Ab(G_{2})}$ by uniqueness. Similarly, $$ \psi\circ\varphi\circ\alpha\circ k_{i}=\psi\circ\ell\circ k_{i}=\psi\circ j_{i}\circ\alpha_{i}=m_{i}\circ\alpha_{i}=\alpha\circ k_{i}, $$ so $\psi\circ\varphi=\Id_{\Ab(G_{1}*G_{2})}$ by uniqueness.

This is just one example - there are many other proofs which seem to follow the same pattern - use the universal properties to derive homomorphisms, compose a bunch of them together, simplify, and prove that you have isomorphisms. My questions are:

  1. Is there a name for this kind of proof?
  2. How can I understand what I'm doing? I feel like I'm just moving symbols around right now, matching up domains/codomains.
  3. Can these proofs be made less tedious?
  4. The reason I say these proofs are like "magic" is that everything seems to fit together perfectly when I'm composing the maps. Why does this happen?
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    $\begingroup$ These are proofs using the "universal characterization" (google universal mapping property) of the groups. Things match up exactly because they have to. There is so little structure in a general category (which is really where your proofs naturally live--abelian category actually) that if things don't work exactly nice, they don't work at all. $\endgroup$ Commented Dec 23, 2012 at 23:58

1 Answer 1


Awesome question. The answer to everything is 'yes.' To elaborate:

As you saw, the key point in all of these exercises is that things satisfy universal properties, and these universal properties give you canonical maps whenever you need them. Another key point (which you've no doubt already noticed) is that a universal property characterizes an object up to isomorphism. Thus a slightly higher-level explanation for the exercise you proved above is that $\mathrm{Ab}(G_1) \oplus (G_2)$ satisfies the universal property characterizing the abelianization of $G_1 * G_2$.

You might call such a proof a 'categorical proof' or 'proof by universal property'; when you're writing out maps and compositions as above, the term 'diagram-chasing' is also frequently used. The advantage of such proofs is that they can often be transported untouched to wholly different settings.

Category theory is essentially the science of doing this, and it sounds like you're at the ideal point to start learning it. Typically when we learn about groups, rings, topological spaces, etc., we think of them as sets of elements with certain structures or operations attached. Category theory subverts this by throwing a curtain over this inner machinery and phrasing everything in terms of the maps which preserve this structure (group homomorphisms, ring homomorphisms, continuous functions of topological spaces, etc.). You probably noticed in your proof above that you didn't need to mention the elements of the groups involved at all -- you only needed to consider the compositions of various homomorphisms between them. This is the basis of categorical thinking. In particular, the very concept of a 'universal property' has a nice, element-free description in category theory.

To give you an example of how nice this can get, using words you may not know yet, an even higher-level explanation of the exercise you proved would be that the free product of groups and the direct sum of abelian groups are both 'coproducts' in their respective categories and the abelianization functor is a 'left adjoint,' which necessarily preserves coproducts. Once you become practiced with this way of thinking, you'll be able to see relationships like this easily, which will make proofs like the one above both more conceptual and less tedious!

A word of warning: it's tempting (at least, it was to me) when you're first learning this stuff to think that it encompasses all of math, and dually to get frustrated with areas of math that don't bend as easily to categorical thinking. No matter how good at it you get, you'll always have to deal with the occasional non-functorial construction or inner goo-style reasoning; in some areas of math, particularly its analytical side, category theory is of very limited utility (though that doesn't stop the categorists from trying to 'explain' these areas too!) It'll certainly change the way you think about a lot of familiar things, though, so if you're of the right mindset for it, you should dive right in.

Start by learning what adjoint functors are and playing around with a lot of examples. I'm partial to Categories for the Working Mathematician by Mac Lane, which only assumes a little bit of experience with algebra, but there are some other good suggestions in this math.SE thread.

  • 1
    $\begingroup$ +1: Your "word of warning" is especially wise - well said! $\endgroup$ Commented Dec 24, 2012 at 1:42
  • $\begingroup$ +1: It’s a good answer, and I especially appreciate if you’re of the right mindset for it, recognizing that ceteris paribus some of us prefer to look at the inner mechanics. $\endgroup$ Commented Dec 24, 2012 at 6:20
  • $\begingroup$ @Paul As an algebraist I love this answer. +1 $\endgroup$
    – user38268
    Commented Dec 26, 2012 at 0:01

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