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Stream: learning: questions

Topic: ✔ Do the tensor products on Ab and CMon have diagonals?

Madeleine Birchfield (Jun 14 2025 at 07:20):

Do the tensor products on [[Ab]] and [[CMon]] have diagonals, in the sense described in [[monoidal category with diagonals]]?

John Baez (Jun 14 2025 at 07:45):

I haven't tried to prove they don't, but it feels obviously wrong. If this were, the subject of algebra would have a different look.

Kevin Carlson (Jun 14 2025 at 07:50):

The obvious guess $a\mapsto a\otimes a$ is not a homomorphism, and it seems fundamental enough to the whole nature of tensor products that you can’t do this that I wouldn’t be very motivated to prove no non-obvious guess works without more reason to suspect the unintuitive.

Madeleine Birchfield (Jun 14 2025 at 08:16):

Toby Bartels claimed that Boolean rings were semilattice objects in Ab on the nLab:

https://ncatlab.org/nlab/revision/diff/Boolean+ring/3

a claim that was just removed earlier today by a Pete Sanders on the nLab

https://ncatlab.org/nlab/revision/diff/Boolean+ring/29

Tobias Fritz (Jun 14 2025 at 08:20):

The only natural diagonal on Ab is to use the zero morphism $A \to A \otimes A$ for every $A$.

In fact, already for $A = \mathbb{Z}$, the only morphism $\mathbb{Z} \to \mathbb{Z} \otimes \mathbb{Z}$ that is natural with respect to endomorphisms of $\mathbb{Z}$ is the zero morphism. That's because these endomorphisms are given by multiplication by a fixed integer $n$, and under $\mathbb{Z} \otimes \mathbb{Z} \cong \mathbb{Z}$, this corresponds to multiplication by $n^2$ on $\mathbb{Z} \otimes \mathbb{Z}$. But the only additive map $f : \mathbb{Z} \to \mathbb{Z}$ that satisfies $f(nx) = n^2 f(x)$ for all $n$ is the zero one.

Madeleine Birchfield (Jun 14 2025 at 08:27):

Is the same true for CMon if we replace $\mathbb{Z}$ in the proof with $\mathbb{N}$?

Tobias Fritz (Jun 14 2025 at 08:30):

Yes, I think that the same argument works for CMon with $\mathbb{N}$ in place of $\mathbb{Z}$.

Notification Bot (Jun 14 2025 at 09:16):

Madeleine Birchfield has marked this topic as resolved.