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I would like to know:

  1. How these two definitions corresponds to each other.

  2. I know that $\otimes$ means the bifunctor in the definition of a monoidal category, thus $I \otimes M$, $M \otimes M$, etc. are all objects in that monoidal category. But what does $\eta \otimes 1$ on the arrow mean? Since $\eta$ is a morphism instead of an object in the monoidal category, I cannot come up with an explanation of its meaning.

Thanks a lot in advance!


1 Answer 1


For 1: In $\mathsf{Set}$, the monoidal product $\otimes$ is actually the Cartesian product $\times$ of sets, and $I$, the unit object, is some one-element set $1 = \{*\}$. Then if $M$ is a monoid, $\mu : M \times M \to M$ is the monoid multiplication, $\eta: 1 \to M$ is the function that sends $*$ to the unit element $e \in M$. The functions $\lambda,\rho$ give the obvious isomorphisms $M \times 1 \cong M \cong 1 \times M$. The first triangle says that $\lambda(*,m) = \mu(\eta(*),\mathrm{id}_{M}(m))$. Just filling in definitions, this gives $m = e \cdot m$. The other triangle similarly gives $m = m \cdot e$.

For 2: A bi-endofunctor on $\mathsf C$ is really just a functor $\otimes: \mathsf C \times \mathsf C \to \mathsf C$. In particular, it should take morphisms of $\mathsf C \times \mathsf C$ to morphisms of $\mathsf C$. Morphisms of $\mathsf C \times \mathsf C$ from $(X,X')$ to $(Y,Y')$ are pairs of morphisms $(f,f')$ where $f: X \to Y$ and $f': X'\to Y'$. Typically, we write $\otimes(f,f')$ as $f \otimes f'$. Thus, $\eta \otimes 1$ is the image of $(\eta,1)$ under the functor $\otimes$. Note that $1$ here refers to $\mathrm{id}_M$.

  • $\begingroup$ I am still a little bit confused about the meaning of $\eta:1->M$, what does it mean to "send * to the unit element $e \in M$? I can understand $\mu$ corresponds to multiplication in Set, since mutiplication can also be seen as a mapping. But how can $\eta$(a mapping) corresponds to the concept of e? Thank you! $\endgroup$
    – Lifu Huang
    Sep 27, 2016 at 11:46
  • $\begingroup$ Recall $1 = \{*\}$, a one-element set. Then $\eta: 1 \to M$ is defined by $\eta(*) = e$. Specifying an element of a set is the same as giving a mapping from $1$ into that set. $\endgroup$ Sep 27, 2016 at 11:48
  • $\begingroup$ I see, thanks. Another question is that: Since $\eta \otimes 1$ is the image of $(\eta, 1)$ under functor $\otimes$, then what's image? It seems that you assumed that $(\otimes(\eta, 1))(\otimes(x, y)) = \otimes(\eta(x), 1(y))$, why does this happen? is there any condition ensuring this? Thanks! $\endgroup$
    – Lifu Huang
    Sep 27, 2016 at 12:08
  • $\begingroup$ A priori, $\otimes(x,y)$ does not mean anything. In the particular case of $\mathsf{Set}$, we have that $\otimes = \times$, and $1 \times \eta$ is the mapping $1 \times M \to M \times M$ that sends $(*,m)$ to $(e,m)$. I.e., it is $\eta$ on the first coordinate and $\id_M$ on the second. This is just the definition of $\times$ as a bifunctor. $\endgroup$ Sep 27, 2016 at 12:15
  • $\begingroup$ That's weird, because from wikipedia, $\otimes$ in Set is just a Cartesian product, but it does not talk anything about how $\otimes$ works on morphisms. Thus I really wonder where I can know this "hidden" definition of $\otimes$ that $otimes(f1, f2)$ in Set means f1 on the first coordinate and f2 on the second? Moreover, In other places, I also often see description of a monoidal category where its $\otimes$ is just (partially?) defined as mapping on objects instead of both on objects and morphisms. Thus I was thinking that there might be some conditions restricting the..(cont'd) $\endgroup$
    – Lifu Huang
    Sep 27, 2016 at 14:31

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