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Stream: theory: category theory

Topic: Natural equalizer preserved by the tensor product in Vec?

Jean-Baptiste Vienney (Jun 02 2024 at 11:50):

Context:

Let $F,G:\mathbf{Vec}_k \rightarrow \mathbf{Vec}_k$ be two endofunctors and $\alpha_A:F(A) \rightarrow G(A)$ a natural transformation. Suppose that for every $A \in \mathbf{Vec}_k$, we have an object $H(A)$ and a morphism $i_A:H(A) \rightarrow F(A)$ such that the following is an equalizer diagram (that is, a kernel diagram):

(This is not very important, but note that in these conditions, we can define $H(f):H(A) \rightarrow H(B)$ for every $f:A \rightarrow B$ and doing so makes $H$ into an endofunctor and $i_A$ into a natural transformation.)

Question:

Are there some equalizer diagrams as above which are preserved by the tensor product, i.e. such that the following is an equalizer digram for every $A \in \mathbf{Vec}_k$?

Comments:

This is not a natural thing to ask for because $\mathbf{Vec}_k$ is monoidal closed and so the tensor product preserves colimits but not limits in general. However, there could be some cases which answer positively the question. I would be interested in nontrivial ones if they exist. A result which says that such a situation almost never happens would be very satisfying as well. An example of a trivial case which answers positively the question is when you choose $\alpha_A = 0$, $H(A)=F(A)$ and $i_A=\mathrm{id}_{F(A)}$.

Todd Trimble (Jun 02 2024 at 12:27):

This should be the case for all vector spaces $A$, because every $A$ is a flat $k$-module (a filtered colimit of finitely generated free modules: consider the union of its finitely generated subspaces). See Theorem 2.5 at [[flat module]].

Jean-Baptiste Vienney (Jun 02 2024 at 12:30):

Awesome! I knew that finite-dimensional vector spaces are flat but I didn't know it is the case for every vector space.

Jean-Baptiste Vienney (Jun 02 2024 at 12:37):

Ok, now I remember. It is even simpler: every free module is flat.

Jean-Baptiste Vienney (Jun 02 2024 at 12:55):

But your proof shows that you don't even need the axiom of choice $\Leftrightarrow$ every vector space is free.