# Definition of unitor in monoidal category

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?

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 $$\mathrm{id}_1 \otimes f : 1 \otimes A \rightarrow 1 \otimes A'$$.

• 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?
– Mark
Feb 20, 2019 at 7:59
• 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 $(1 \otimes (-))(f) := h \otimes f$. Feb 20, 2019 at 9:22
• 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.
– Mark
Feb 20, 2019 at 17:02
• @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 $\otimes$. Feb 20, 2019 at 19:47
• @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.
– Mark
Feb 21, 2019 at 5:40