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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:

  1. 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.
  2. 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.
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    $\begingroup$ A simple example of non-preservation: Take the signature with two constant symbols $a$, $b$ and $T$ the theory with no equations and $T'$ the theory consisting of $a = b$. The left adjoint sends the $T$-model $S = \{a \ne b\}$ to a one-element set but $S \times S$ to a three-element set. $\endgroup$ – Reid Barton Mar 11 at 12:46
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    $\begingroup$ I have no general answer, but for a simple example where product-preservation 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 12:48
  • $\begingroup$ Thanks for the minimalistic counterexamples! So the idea can be nicely summarized as orbits not preserving products. Take the theory of $G$-sets and of "trivial" $G$-sets, then taking orbits is left adjoint to the inclusion. $\endgroup$ – Dario Stein Mar 11 at 19:41
  • $\begingroup$ I'm wondering if the situation changes once we have nontrivial $n$-ary operations for $n$>1. E.g. colimits only really start being messy because $n$-ary operations intermix elements from different copies. $\endgroup$ – Dario Stein Mar 11 at 19:44
  • $\begingroup$ Actually nevermind, the same counterexample can be extended to binary operations: Take the theory of magmas, and add the extra "trivialization" equation $x\cdot y = x$. Then the left adjoint does not preserve the product $2 \times 2$ where we define the magma structure on $2$ by $x \cdot y = \neg x$. $\endgroup$ – Dario Stein Mar 12 at 11:53
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The coproduct of groups $A$ and $B$ (which I think group theorists call amalgamation) consists of alternating strings of non-identity elements of $A$ and $B$ of any finite length. The empty string is the identity and the new group product is given by concatenation, with appropriate cancellation.

This construction is completely familiar for "fundamental" groups of homotopies and is the basis of van Kampen's theorem.

The product is the same, except that elements of $A$ commute with those of $B$. This means that the strings reduce to pairs.

For Abelian groups, everything commutes with everything else, so that the product and coproduct of Abelian groups are the same.

In the Question, the individual terms of the strings or pairs are Abelianised, ie when several elements of $A$ are combined they commute, and similarly $B$.

These two processes - Abelianising amongst terms for the product and within them for Abelianising individual groups - clearly commute.

As a constructive mathematician, I consider it sinful to distinguish non-identity elements, but there is a more complicated constructive argument using quotients by equivalence relations.

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    $\begingroup$ Group theorists call the coproduct the free product, which in turn is distinct from (what you might be thinking of) a free product with amalgamation. $\endgroup$ – Carl-Fredrik Nyberg Brodda Mar 11 at 12:01
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    $\begingroup$ Sorry, I only read half of this long question and thought I could give a quick answer just before logging off and going out. The more general question is more interesting and not obviously one for a categorist. I wonder whether my argument might give a hint. $\endgroup$ – Paul Taylor Mar 11 at 17:46
  • $\begingroup$ Thank you Paul, for the special case of abelianization you gave a great intuition. Though you seems to be using the trick that for groups, the product is actually a quotient of the coproduct. This is strange categorically. In order to even define $G * H \to G \times H$ one needs unitality/zero objects again. $\endgroup$ – Dario Stein Mar 11 at 19:14
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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:

  1. $B: Grp \to sSet$ 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.

  2. $F_{\mathbb Z}: sSet \to sAb$ 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:

  1. 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.

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