From https://ncatlab.org/nlab/show/monoidal+category, a monoidal category requires a natural isomorphism

  1. a natural isomorphism $\lambda: (1 \otimes (-)) \rightarrow ^\cong (-)$ with components of the form $\lambda_x : 1 \otimes x \rightarrow x$

What does $(-)$ mean? Maybe the identity endofunctor? I suppose $(1 \otimes (-))$ is shorthand for a endofunctor which operates on objects by multiplying them by 1, but how does it operate on arrows?


1 Answer 1


Let $\mathscr{A}$ be a monoidal category. Let $A,A' \in \mathscr{A}$, and suppose $f \in \mathscr{A}(A,A')$.

You are right that $(-): \mathscr{A} \rightarrow \mathscr{A}$ is $\mathrm{id}_{\mathscr{A}}$.

$(1 \otimes (-)) : \mathscr{A} \rightarrow \mathscr{A}$ is defined as:

action on object $A$ given by $1 \otimes A$,

action on morphism $f$ given by $\mathrm{id}_1 \otimes f : 1 \otimes A \rightarrow 1 \otimes A'$.

  • $\begingroup$ That makes sense. Is the action on morphisms fully constrained by the previous 3 requirements, or is the ncat link missing this clarification, or is it convention? $\endgroup$
    – Mark
    Commented Feb 20, 2019 at 7:59
  • $\begingroup$ Given that $(1 \otimes (-))(A) = 1 \otimes A$ and $(1 \otimes (-))(A') = 1 \otimes A'$ we know that $(1 \otimes (-))(f) : 1 \otimes A \rightarrow 1 \otimes A'$, so choosing $\mathrm{id}_1 \otimes f$ is the obvious (and therefore correct) choice. Besides, the only morphism $1 \rightarrow 1$ that you are guaranteed to have exist is $\mathrm{id}_1 : 1 \rightarrow 1$, so it's not like you even have the choice to fix some $h: 1 \rightarrow 1$ and take $(1 \otimes (-))(f) := h \otimes f$. $\endgroup$ Commented Feb 20, 2019 at 9:22
  • $\begingroup$ I think it might also be impossible to construct a perverse monoidal category which makes another choice because of the consistency conditions but I'm not sure. $\endgroup$
    – Mark
    Commented Feb 20, 2019 at 17:02
  • $\begingroup$ @Mark The fact that $1\otimes (-) (f)=\mathrm{id}_1\otimes f$ is just how restricting a functor of two variables to a one-variable functor works. There's no opportunity to make any choices here once you have the bifunctor $\otimes$. $\endgroup$ Commented Feb 20, 2019 at 19:47
  • $\begingroup$ @KevinCarlson The bifunctor $\otimes$ is a functor between $C \times C \rightarrow C$. So it would seem that valid expressions involving $\otimes$ are $a \otimes b$ where $a$ and $b$ are both objects or both morphisms where the result is an object or morphism of $C$. But here we have the expression $1 \otimes (-)$ so $a = 1$ is an object and $b = (-)$ is a functor, and the result is a functor. I guess I'm confused on what are the valid rules of functor manipulations like this. $\endgroup$
    – Mark
    Commented Feb 21, 2019 at 5:40

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