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I found two definitions of 'strict monoidal category'.

First definition: a strict monoidal category is a monoid in $\operatorname{Cat}$.

Definition 4.9. A strict monoidal category is a category $\mathbf{C}$ equipped with a binary operation $\otimes: \mathbf{C}\times\mathbf{C}\to\mathbf{C}$ which is functorial and associative, $$A\otimes(B\otimes C)=(A\otimes B)\otimes C,\qquad\qquad (4.2)$$ together with a distinguished object $I$ that acts as a unit, $$I\otimes C=C=C\otimes I.\qquad\qquad\qquad\qquad (4.3)$$

My question is: are the two definitions equivalent? In particular, how can I deduce the associativity of $\otimes$ on morphisms, which is included in the first definition, from the associativity of $\otimes$ on objects, that is the only one I have in the second definition?

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  • $\begingroup$ I found the second definition on Awodey, "Category theory". It means that $1\otimes A=A=A\otimes 1$ for each object $A$ of $\mathcal{C}$. $\endgroup$ – bateman May 17 '18 at 13:08
  • $\begingroup$ Sorry, I wrote naturally in place of neutrally, I am editing that. $\endgroup$ – bateman May 17 '18 at 13:09
  • $\begingroup$ @Hurkyl I copy-pasted the Awodey definition in the text of the question. $\endgroup$ – bateman May 17 '18 at 13:18
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Presumably, the given equations are meant to be of the indicated functors, not just the object mappings. Identity 4.2 is meant to say that $\otimes$ is associative both on objects and on morphisms.


If you only require $\otimes$ to be associative on objects, then these two definitions would indeed be different. For example, let $A$ be an abelian group, and $\varphi$ be a nontrivial automorphism of $A$.

Let $\mathcal{A}$ be the one-object category corresponding to $A$. We can make $\mathcal{A}$ a category satisfying the "objects only" interpretation of definition 4.9 by defining it on morphisms by the identity

$$ a \otimes b = \varphi(ab) $$

It's easy to check that this is a bifunctor:

  • $1 \otimes 1 = \varphi(1 1) = 1$
  • $(a \otimes b)(c \otimes d) = \varphi(ab) \varphi(cd) = \varphi(abcd) = \varphi(ac) \varphi(bd) = (ac) \otimes (bd)$

and since $\mathcal{A}$ only has one object, it's trivial to satisfy the two equational identities.

However, this operation need not be associative on morphisms, since

$$ a \otimes (b \otimes c) = \varphi(a \varphi(bc)) = \varphi(a) \varphi^2(b) \varphi^2(c)$$ $$ (a \otimes b) \otimes c = \varphi(\varphi(ab) c) = \varphi(a)^2 \varphi^2(b) \varphi(c)$$

and thus associativity is equivalent to the assertion that

$$ a \varphi(c) = \varphi(a) c$$ for every $a,c \in A$.

If, for example, we chose the automorphism $\varphi(x) = x^{-1}$, then this equation becomes $a^2 = c^2$, and its easy to find groups that don't satisfy this identity.

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  • $\begingroup$ I do agree, I think that $A\otimes (B\otimes C)$ is meant to be $(-)\otimes ( (-)\otimes (-))$. $\endgroup$ – bateman May 17 '18 at 14:48

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