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Stream: theory: category theory

Topic: monoidal products do not preserve monos

Robin Piedeleu (Nov 16 2021 at 09:38):

Embarrassingly, I've realised I don't know a good counterexample to the claim that monoidal products preserve monos, i.e. that the monoidal product of two monos is still mono. There's no reason why it should be the case, so what's a counterexample?

We need to look for a counterexample in a category where monos don't split. It is enough to find/construct (depending on your philosophical bend) a monoidal category with a monomorphism $m : A\rightarrow B$ and two different morphisms $f,g : I \rightarrow A\otimes C$ satisfying $(m\otimes id_C)\circ f = (m\otimes id_C)\circ g$.

Graham Manuell (Nov 16 2021 at 13:44):

Consider the inclusion $m\colon \mathbb{Z} \to \mathbb{Q}$ in the category of abelian groups. Tensoring with the identity on $\mathbb{Z}/2\mathbb{Z}$ gives the unique map $\mathbb{Z}/2\mathbb{Z} \to 0$ which is certainly not monic.

John Baez (Nov 16 2021 at 14:42):

How about the category of abelian groups with its usual tensor product? Consider the morphism "multiplication by $n$"

$n : \mathbb{Z} \to \mathbb{Z}$

and the identity morphism

$1 : \mathbb{Z}/n \to \mathbb{Z}/n$

These are both monos, but if we tensor them we get

$0 : \mathbb{Z}/n \to \mathbb{Z}/n$

John Baez (Nov 16 2021 at 14:44):

Oh, darn. I tend to reply to posts before seeing if someone else already did. Anyway, yeah, tensoring in categories of $R$-modules is rather famously not "left exact", so it doesn't preserve monos. The $\mathrm{Tor}$ functors arise from this fact.

Robin Piedeleu (Nov 16 2021 at 15:48):

Thank you to both! (I figured I should have looked into categories of modules but my commutative algebra was too rusty.)

John Baez (Nov 16 2021 at 15:53):

Yeah, when you learn algebraic topology, like homology theory, they hit you over head with how tensoring by an object doesn't preserve exact sequences: it preserves epis but not monos. To deal with that you have to learn about Tor, and then derived functors... and so on. So I was glad to see a question I'd been taught the answer to, long ago.

Jens Hemelaer (Nov 16 2021 at 20:45):

Interesting question and nice counterexample. Does someone know if a counterexample exists such that the domain of the two morphisms $f$ and $g$ is the same?

Reid Barton (Nov 16 2021 at 22:11):

There's a monomorphism $\mathbb{Z}/2 \to \mathbb{Q}/\mathbb{Z}$ (that sends the nontrivial element to $1/2$) but $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$!

Reid Barton (Nov 16 2021 at 22:14):

I always find this example counterintuitive, based on some half-baked idea that we could reduce to working over one field at a time, where tensoring something with itself can't make it 0.

Reid Barton (Nov 16 2021 at 22:15):

Concretely, we have $(1/2) \otimes (1/2) = (1/4 + 1/4) \otimes (1/2) = (1/4) \otimes (1/2) + (1/4) \otimes (1/2) = (1/4) \otimes 1 = 0$.

Reid Barton (Nov 16 2021 at 22:17):

I guess for the question about tensoring monos we can also take the map $\mathbb{Z}/2 \to \mathbb{Z}/4$ sending 1 to 2.

John Baez (Nov 16 2021 at 22:18):

Hey, that's nice. I was thinking about that $\mathbb{Z}/4$ example for a minute but it didn't seem to be working.

Jens Hemelaer (Nov 17 2021 at 07:47):

Reid Barton said:

There's a monomorphism $\mathbb{Z}/2 \to \mathbb{Q}/\mathbb{Z}$ (that sends the nontrivial element to $1/2$) but $\mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$!

Wonderful example, thanks!