Is this relationship on subcartesian products of quotient groups over the free group correct?

Question

I am trying to check if the following relationship is correct. Let $F$ be a free group over some set $X$. Let $\alpha_1, \alpha_2, ..., \alpha_n$ be epimorphisms from $F$ to groups $H_1, H_2, ..., H_n$ respectively. Let $K_1, K_2, ..., K_n$ be the kernels of $\alpha_1, \alpha_2, ..., \alpha_n$ respectively. It follows that $K_1, K_2, ..., K_n$ are all normal.

Now, let $K= \bigcap K_i$ for $i=1,...,n$ (i.e. the normal group formed from the intersection of the kernels). We can now form the quotient $F/K$.

Is it true that $F/K$ is isomorphic to a subcartesian product of $F/K_1,F/K_2,...,F/K_n$ ? How could I show it ?

I want to check if this is correct since this would then mean that $F/K$ is also isomorphic to a subcartesian product of $H_1,H_2,...,H_n$ ...

(For context over this question, am trying to understand a proof for Birkhoff's variety theorem which uses the above)

Answer 1

Edit: as pointed out in the comments, "subcartesian product" has a formal meaning and is not the same as a cartesian product of subgroups.

Some thoughts which I found too long for a comment:

The maps $\alpha_i$ define a map

$$ f \colon t \in F \mapsto (\alpha_1(t),\ldots, \alpha_n(t)) \in H_1 \times \cdots \times H_n $$

Its kernel consists of the $t \in F$ for which $\alpha_i(t) = 1$ for all $i$ simultaneously, i.e. $K = \cap_i K_i$. Hence

$$ F/K \simeq \mathbf{im} f \leq H_1 \times \cdots \times H_n. $$

We have not used that each $\alpha_i$ is epi nor the fact that $F$ is free. In general, the image of $f$ need not be a cartesian product of subgroups, even with these hypotheses (e.g. if $F = \Bbb Z$, $n = 2$, and $\alpha_1 = \alpha_2 = id$, then $f$ is the diagonal map $\Delta \colon \Bbb Z \to \Bbb Z^2$).

It may be that $F/K$ is still isomorphic to some cartesian product through some other morphism.

Answer 2

Recall that a subdirect product of $\{G_i\}_{i\in I}$ is a subgroup $M\leq \prod_{i\in I}G_i$ such that $\pi_j(M)=G_j$ for all $j\in I$.

In general, if $G$ is a group, $\{H_i\}_{i\in I}$ is a family of groups, and $f_i\colon G\to H_i$ are morphisms, then we obtain an induced homomorphism $$f\colon G\to \prod_{i\in I}H_i$$ by the universal property of the product. This map has the property that for each $i\in I$, $\pi_i\circ f = f_i$. If $I=\{1,\ldots,n\}$, you can "visualize" $f$ as the map $f(g) = (f_1(g),\ldots,f_n(g))$.

In this case, the image of $f$ is always a subdirect product of the family $\{f_i(G)\}_{i\in I}$.

Indeed, given $h_i\in f_i(G)$, we know there exists $g\in G$ such that $f_i(g)=h_i$. Then $\pi_i(f(g)) = f_i(g) = h_i$ by definition, so there exists an element in $\mathrm{Im}(f)$ that has $i$th coordinate equal to $h_i$. This proves that $f(G)$ is a subdirect product of $\{f_i(G)\}_{i\in I} = \{\mathrm{Im}(f_i)\}_{i\in I}$.

In your situation, $G=F$ is a free group and the maps $\alpha_i$ are surjective (I avoid "epimorphism" because there are varieties of groups where you have nonsurjective epimorphisms; of course, in the category of all groups epimorphisms are surjective, but I prefer not to conflate the two). So certainly the map $F\to H_1\times\cdots\times H_n$ has image that is a subdirect product. The only thing remaining is to verify that the morphism factors through $K$, but that follows because the kernel of the universal map $\alpha\colon F\to H_1\times\cdots\times H_n$ is precisely the intersection of the kernels of the maps $\alpha_i$, which is $K$.