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The associativity condition in a strict monoidal category is very often defined just at the object level, namely, for objects $x,y,z$ in a strict monoidal category $\mathsf{C}$, one has:

$$(x\otimes y)\otimes z = x\otimes (y\otimes z)$$

Does the associativity at morphism level derive from this condition or is it a part of the definition. As I see, for $f:x\to x'$, $g:y\to y'$ and $h:z\to z'$ the two morphisms $(f\otimes g)\otimes h$ and $f \otimes (g\otimes h)$ share the same source and target objects but are not necessarily the same (what really does the same mean in strict case!).

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Yes, they are equal. This follows from the naturality of associator, which is identity in strict monoidal categories.

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    $\begingroup$ Equivalently (I think), strict associativity means not only the equation displayed in the question but also that the associativity isomorphism between the two (equal) sides of that equation is not only an isomorphism but the identity. $\endgroup$
    – Andreas Blass
    Jul 30, 2016 at 15:59
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    $\begingroup$ It's true that they are equal in a strict monoidal category, but I think the answer to the question as stated is "no", because this doesn't follow from asserting a simple equality on objects; that equality has to be a "natural equality". $\endgroup$
    – Mike Shulman
    Jul 31, 2016 at 7:01
  • $\begingroup$ @AndreasBlass, "the associativity isomorphism between the two (equal) sides of that equation is not only an isomorphism but the identity" this is the difference between a strict monoidal category and a non-strict one, in a non strict one the two objects are merely isomorphic, in a strict one they are equal. Now, I heard someday that isomorphism IS equality... $\endgroup$
    – Pedro
    Jul 31, 2016 at 8:14
  • $\begingroup$ @MikeShulman what do you mean by "natural equality"? $\endgroup$
    – Pedro
    Jul 31, 2016 at 8:15
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    $\begingroup$ It is a part of the definition. It requires not only objects to be equal but also associator to be identity. Equality of morphisms follows from the second condition. $\endgroup$
    – Valery Isaev
    Jul 31, 2016 at 8:56

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