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.

fails: take $T$ to be the empty theory over a single unary operation; take $T'$ to add the axiom that it’s the identity. Write $F$ for the left adjoint “$T'$-isation”, and take $A = \mathbb{Z}/2$ with its involution. Then $F(A)$ is a singleton, but $F(A \times A)$ has two elements, so $F(A \times A) \not \cong F(A) \times F(A)$. $\endgroup$ – Peter LeFanu Lumsdaine Mar 11 at 12:482more comments