TITLE: Why does abelianization preserve finite products, really?
QUESTION [19 upvotes]: The abelianization functor $(-)^{ab} : \mathrm{Grp} \to \mathrm{Ab}$ is left adjoint to the inclusion of abelian groups into groups. As such, it preserves all colimits, but it doesn't generally preserve limits (e.g. the mono $A_3 \hookrightarrow S_3$ is not preserved). Now curiously, it does preserve finite products, that is $$(G \times H)^{ab} \cong G^{ab} \times H^{ab}$$ I am trying to find an intuition why this is true from an abstract point of view: Which special properties of (abelian) groups really go into proving this?
Before I generalize, let's recall one possible proof of product preservation:

The equivalence relation for the abelianization of $G$ can be
described as follows: Let $g \sim g'$ if for all homomorphisms $\phi : G \to A$ into abelian groups, we have $\phi(g) = \phi(g')$. Write
$[g]$ for equivalence classes. This is a congruence and $G/\!\sim$ has
the universal property of the abelianization. To show that the
comparison morphism $(G \times H)^{ab} \to G^{ab} \times H^{ab}$ given
by $[(g,h)] \mapsto ([g],[h])$ is an iso, we attempt to show that the
obvious inverse $([g],[h]) \mapsto [(g,h)]$ is well-defined: That is,
if $g \sim g', h \sim h'$ then $(g,h) \sim (g',h')$. Let $\phi : G
 \times H \to A$ be a homomorphism into an abelian group, then
$\phi(-,1), \phi(1,-)$ are homomorphisms separately. Hence as desired
$$\phi(g,h) = \phi((g,1)\cdot(1,h)) = \phi(g,1) \cdot \phi(1,h) =
 \phi(g',1)\cdot\phi(1,h') = \phi(g',h')$$

Note I had to use the neutral element as a way of relating homomorphisms out of a product to homomorphisms of the factors.
Towards a generalization, let us look at a different example: A convex sets is a set equipped with convex-combination operations $x +_r y$ for $0 \leq r \leq 1$. A semilattice is a convex set where $+_{1/2}$ is associative. This forces all operations $+_r$ for $0 < r < 1$ to become equal, associative and a semilattice operation, jointly written $\vee$. Like abelianization, the inclusion of semilattices into convex sets has a left adjoint $(-)^{col} : \mathrm{Cx} \to \mathrm{Sl}$ which "collapses" probability to possibility so to speak, identifying all points which lie on the interior of some line segment. Again, this operation preserves products, but now for a slightly different reason: We don't have a neutral element, however for $\phi : X \times Y \to Z$ and fixed $y$, $\phi(-,y)$ is a homomorphism because all operations are idempotent ($y+_r y=y$).
I arrive at my general question:

Fix some signature. Let $T$ be an algebraic theory and $T' \supset T$ a super-theory over the same signature. The inclusion of $T'$-algebras into $T$-algebras has a left adjoint, freely enforcing the additional $T'$-equations. When does this left adjoint preserve products?

From the above examples this seems to require particular properties of the respective theories. Yet, I don't know an example where product preservation does not hold, and would appreciate one if it exists. I'd also like other (category-theoretic) insights into the abelianization case.
Possible lines of thought:

The result doesn't seem to be a straightforward instance of the fact that reflexive coequalizers commute with finite products, though certainly coequalizers are involved.
This proof to the abelianization case constructs the inverse $G^{ab} \times H^{ab} \to (G \times H)^{ab}$ by using the fact that in $\mathrm{Ab}$, the product $G^{ab} \times H^{ab}$ is actually a biproduct, by which it is enough to find morphisms $G^{ab},H^{ab} \to (G \times H)^{ab}$ separately. Again, this situation seems rather particular.

REPLY [3 votes]: Here's an answer which doesn't really seem to lend itself to thinking about general algebraic theories, but does put this result in some larger context: think about it topologically!

Review of the Hurewicz theorem for $n=1$:
Consider the Hurewicz theorem for $n=1$, which says that for any based space $X$, we have $H_1(X) = \pi_1(X)^{ab}$. Working simplicially or cellularly, this case of the theorem is totally elementary to prove: given $X$, the van Kampen theorem implies that a presentation of its fundamental group is given by taking one generator for each nondegenerate 1-cell in the connected component of the basepoint, and a relation is imposed for each nondegenerate 2-cell between these. On the other hand, $H_1(X)$ is the first homology of the free simplicial abelian group on the simplicial set $X$, i.e. it has one generator for each 1-cell and a relation $d_1(x) = d_2(x) + d_0(x)$ for each 2-cell. That is, it is the free abelian group with the same presentation. Since abelianization is a left adjoint, it preserves the coproducts and coequalizers implicit in saying that these are "presentations", and so it carries the former to the latter.

Now I claim that from this description, we can read off the fact that abelianization preserves finite products. For we are saying that if $G$ is a group, then $G^{ab} = H_1(BG) = \pi_1(F_{\mathbb Z}(BG))$. So we've broken $(-)^{ab}$ down as a composite of 3 functors:

$B: Grp \to sSet$ really factors as the composite $B = N\mathbb B$, where $\mathbb B: Grp \to Cat$ carries a group $G$ to $\mathbb B G$, i.e. the same group considered as a 1-object category, and $N: Cat \to sSet$ is the nerve functor. Both of these functors are right adjoints, and in particular preserve finite products.

$F_{\mathbb Z}: sSet \to sAb$ is the functor which carries a simplicial set $X$ to the free abelian group $F_{\mathbb Z}X$ on $X$ (viewing $X$ as a functor $\Delta^{op} \to Set$, this is just postcomposing with the free abelian group functor $Set \to Ab$). Since the free abelian group functor is a strong symmetric monoidal functor $(Set,\times) \to (Ab,\otimes)$, it follows that $F_{\mathbb Z}$ is a strong symmetric monoidal functor $sSet \to sAb$.


At this point, one should worry a little bit -- the composite $F_{\mathbb Z} B: Grp \to sAb$ is strong symmetric monoidal, but with respect to a different monoidal structure on $sAb$ than the cartesian one ($\otimes \neq \times$). Never fear:

The functor $\pi_1: sAb \to Ab$ doesn't carry tensor products to cartesian products in general. However, when $A = F_{\mathbb Z} X$, $B = F_{\mathbb Z} Y$ are the free simplicial abelian groups on simplicial sets $X,Y$ which each have a unique 0-cell, we get that $(A \otimes B)_0 = \mathbb Z$ and $(A\otimes B)_1 = A_0 \otimes B_1 \oplus A_1 \otimes B_0 = B_1 \oplus A_1$. Similarly, we have $(A\otimes B)_2 = A_0 \otimes B_2 \oplus A_1 \otimes B_1 \oplus A_2 \otimes B_0 = B_2 \oplus A_1 \otimes B_1 \oplus A_2$; differentials from the first and third summands enforce relations on $B_1,A_1$ to cut down to $H_1(B),H_1(A)$, while differentials from the middle term ensure that sums from the two factors commute with each other.

The picture I want to get across is that very roughly (this is actually true over a field) we have $H_n(X \times Y) = \oplus_{i+j} = H_i(X) \otimes H_j(Y)$; when $n=1$ the only terms we have to pick up are $H_1(X) \otimes H_0(Y)$ and $H_0(X) \otimes H_1(Y)$, whose sum gives the desired cartesian product.