Intersection of subalgebras in an abelian monoidal category

Question

$\require{AMScd}$Let $\mathcal{C}$ be an abelian monoidal category, and let $(M,m,e)$ be an algebra in $\mathcal{C}$, where

• $M$ is an object in $\mathcal{C}$,
• $m: M \otimes M \to M$ is the multiplication morphism,
• $e: 1\to M$ is the unit morphism,

satisfying the associativity and unit axioms.

A subalgebra $X$ of $M$ is defined as an algebra $(X, m_X, e_X)$ together with an monomorphism $i_X: X \to M$ which is an algebra morphism, i.e.

• $e = i_X \circ e_X$,
• $i_X \circ m_X = m \circ (i_X \otimes i_X)$.

Let $A$ and $B$ be two subalgebras of $M$. According to [Fr64, §1.5], the intersection $A \cap B$ can be described by the following pullback diagram: $$ \begin{CD} A \cap B @>j_A>> A \\ @VVj_BV @VVi_AV \\ B @>i_B>> M \end{CD} $$

Question: Is there an algebra structure on $A \cap B$ such that $i_A \circ j_A = i_B \circ j_B$ is an algebra monomorphism?

Roughly speaking, we are asking whether the intersection of two subalgebras is still a subalgebra. This is obviously true for $\mathcal{C} = \text{Vec}$, but it may not hold in general (although I do not have a counterexample at the moment). If so, what happens when we restrict to a tensor category $\mathcal{C}$ over $\mathbb{C}$? If it also does not hold in this case, what would be a minimal relevant assumption on $\mathcal{C}$?

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Reference:

[Fr64] Freyd, Peter. Abelian Categories: An Introduction to the Theory of Functors. Harper's Series in Modern Mathematics. Harper & Row, Publishers, New York, 1964. xi+164 pp.

Answer 1

Yes, such an algebra structure exists. In fact, there is a unique algebra structure on $A \cap B$ such that $j_A : A \cap B \to A$ and $j_B : A \cap B \to B$ are algebra morphisms.

To construct the morphisms $m_{A \cap B} : (A \cap B) \otimes (A \cap B) \to A \cap B$ and $e_{A \cap B} : 1 \to A \cap B$ and check the necessary axioms, one uses the universal property of pullbacks.

For example, to construct $m_{A \cap B}$, we can use $$m_A \circ (j_A \otimes j_A) : (A \cap B) \otimes (A \cap B) \to A$$ and $$m_B \circ (j_B \otimes j_B) : (A \cap B) \otimes (A \cap B) \to B.$$ One checks that $$ i_A \circ m_A \circ (j_A \otimes j_A) = i_B \circ m_B \circ (j_B \otimes j_B)$$ and so there is a unique map $m_{A \cap B} : (A \cap B) \otimes (A \cap B) \to A \cap B$ such that $$j_A \circ m_{A \cap B} = m_A \circ (j_A \otimes j_A)$$ and $$j_B \circ m_{A \cap B} = m_B \circ (j_B \otimes j_B).$$ The following diagram summarizes the conclusion. Note that this includes part of the information needed to show that $j_A$ and $j_B$ are algebra morphisms. The rest of the argument is similar.

If things are set up correctly, one can actually use the same exact argument to show that the functor $\mathrm{Alg}(\mathcal{C}) \to \mathcal{C}$ forgetting algebra structures creates limits. This only requires $\mathcal{C}$ to be a monoidal category.