$\require{AMScd}$ Let $\mathcal{C}$ be a tensor category. An object $X$ in $\mathcal{C}$ will be called *selfdual* if $X^* = X$. Let $A,B,M$ be selfdual objects in $\mathcal{C}$. Let $i_X: X \to M$, with $X=A,B$, be two monomorphisms, and consider their pullback below:

$$\begin{CD} P @>j_A>> A\\ @VVj_BV @VVi_AV\\ B @>i_B>> M \end{CD}$$

**Question**: Is the object $P$ also selfdual (up to isomorphism)?

Following Freyd's book [Fr64, pages 19, 37-40], $i_X$ is a (equivalent class representative) subobject of $M$, and $P$ is the intersection $A \cap B$. If we can prove something like $(A \cap B)^* \simeq A^* \cap B^*$, then we are done.

If required, we can assume, for $X=A,B$, that $i_X^* \circ i_X = {\rm id}_X$, implying that $i_X$ is a *split* monomorphism. Thus, by splitting lemma, we can assume without loss of generality that $$M = X \oplus M/X$$ with $i_X = {\rm id}_X \oplus 0$ (recall that $X = X \oplus 0$).

The dual of above diagram is the following pushout of epimorphisms $i_X^*: M \to X$.

$$\begin{CD} M @>i_A^*>> A\\ @VVi_B^*V @VVj_A^*V\\ B @>j_B^*>> P^* \end{CD}$$

**Semisimple case**: Let $(X_i)$ be the simple objects up to isomorphism. Without loss of generality, we can take $M=\bigoplus_i M_i \otimes X_i$, $A=\bigoplus_i A_i \otimes X_i$ and $B=\bigoplus_i B_i \otimes X_i$, where $A_i$ and $B_i$ are subspaces of the multiplicity space $M_i$, for all $i$. Without loss of generality, we can take $i_A$ and $i_B$ induced by the inclusions $A_i, B_i \subset M_i$. Then $P=\bigoplus_i P_i \otimes X_i$, with $P_i = A_i \cap B_i$. But $A$, $B$, $M$ are selfdual, so $M_{i^*} = M_i$, $A_{i^*} = A_i$ and $B_{i^*} = B_i$. Thus $$P_{i^*} = (A_i \cap B_i)^* = A_{i^*} \cap B_{i^*} = A_i \cap B_i = P_i,$$ for all $i$, meaning that $P^* = P$.

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. {\rm xi}+164 pp.