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Let $(C,\otimes,1)$ be a braided monoidal category with braiding $\beta$. Given two monoidal objects, $A$ and $B$, with multiplication maps $\mu_A$ and $\mu_B$ and unit maps $e_A$ and $e_B$, I've read that $A \otimes B$ is a monoidal object with multiplication given by $(\mu_B \otimes \mu_A) \circ (1 \otimes \beta \otimes 1)$ and unit $e_A \otimes e_B$. I'm having trouble showing that this is true. I couldn't find the proof in Maclane or by a google search. Does anyone know how to show this?

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Ok I figured it out. I was forgetting to use naturality of $\beta$. This would be easier to write if I knew how to code diagrams in Latex, but I'll give it a shot. Let $\beta_{A,B}$ denote the braiding $A \otimes B \rightarrow B \otimes A$. By naturality of $\beta$ we have that for any morphisms $f : A \rightarrow C$ and $g : B\rightarrow D$, we have $\beta_{A,B} \circ (f\otimes g) = (g\otimes f) \circ \beta_{C,D}$.

Let $\mu = (\mu_A \otimes \mu_B) \circ (\text{id}_A\otimes \beta_{B,A} \otimes \text{id}_B)$.

For associativity we want to show that $$\mu\circ (\mu\otimes \text{id}_{A\otimes B}))= \mu \circ (\text{id}_{A\otimes B}\otimes\mu)$$

To do this, show both sides equal $(\mu_A ^2 \otimes \mu_B ^2) \circ \beta$ where $\beta$ is the braiding $A\otimes B \otimes A\otimes B \otimes A\otimes B \rightarrow A\otimes A \otimes A\otimes B \otimes B\otimes B$ and $\mu_A ^2$ and $\mu_B ^2$ are well defined by associativity.

We have:

$\mu\circ (\mu\otimes \text{id}_{A\otimes B})$

$= (\mu_A \otimes \mu_B) \circ (\text{id}_{A} \otimes \beta_{B,A} \otimes \text{id}_{B})\circ ([(\mu_A \otimes \mu_B) \circ (\text{id}_{A}\otimes \beta_{B,A} \otimes \text{id}_{B})]\otimes \text{id}_{A\otimes B} )$

$= (\mu_A \otimes \mu_B) \circ (\text{id}_{A} \otimes \beta_{B,A} \otimes \text{id}_{B})\circ (\mu_A \otimes \mu_B\otimes \text{id}_{A\otimes B} ) \circ (\text{id}_{A}\otimes \beta_{B,A} \otimes \text{id}_{B} \otimes \text{id}_{A\otimes B} )$ (Since $\otimes$ is a functor)

$= (\mu_A \otimes \mu_B) \circ (\mu_A \otimes \text{id}_{A} \otimes \mu_B\otimes \text{id}_{B}) \circ (\text{id}_{A\otimes A} \otimes \beta_{B\otimes B,A} \otimes \text{id}_{B})\circ (\text{id}_{A}\otimes \beta_{B,A} \otimes \text{id}_{B}\otimes \text{id}_{A\otimes B})$ (By naturality)

$=(\mu_A ^2 \otimes \mu_B ^2) \circ \beta$

Do the analogous thing for the other side. Note that we use associativity of $\mu_A$ and $\mu_B$ to see that these expressions are equal.

I'll do the unit map later.

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