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Let $(M,\otimes)$ be a rigid monoidal category, for which left and right duals coincide. For any object $X \in M$, we can define a monoid structure on $X \otimes X^*$: Multiplication is defined by evaluation $ev$ as follows: $$ id \otimes ev \otimes id: X \otimes X^* \otimes X \otimes X^* \to X \otimes X^*. $$ The unit is just given by coevaluation in the obvious way.

Is this definition correct? I can convince myself that the unit axiom of a monoid object is true - it follows from the axioms of a dual. But I'm confused about the associativity axiom.

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    $\begingroup$ Yes, it's correct. Associativity is easiest to prove using string diagrams; a bicategorical version of this argument shows that an adjunction gives rise to a monad (and in particular you don't need an assumption about left vs. right duals). The monoid structure you get this way is an enriched endomorphism object of $X$ with respect to a (partial) enrichment of $M$ over itself, generalizing the familiar case of $\text{FinVect}$. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 3, 2021 at 20:45
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    $\begingroup$ $X\otimes X^*$ is also known as the internal hom of $X$; and you can prove more generally that $\hom(Y,Z)\otimes \hom(X,Y)\to \hom(X,Z)$ is "associative" in a monoidal closed category - this has appeared somewhere on MO or MSE - more specifically here : mathoverflow.net/questions/21382/… (you then need to compare your map on $X^*\otimes X$ to the one on $\hom(X,X)$) $\endgroup$
    – Maxime Ramzi
    Commented Jan 3, 2021 at 20:50
  • $\begingroup$ See Lemma 4.11 and further in global.oup.com/academic/product/… $\endgroup$
    – Chris Heunen
    Commented Jan 4, 2021 at 13:56

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