In Wikipedia we can find the following "Pentagon Diagram" for Monoidal Categories.

enter image description here

What I don't understand is why that diagram is constructed in that way.

Why do we start from $((A \otimes B) \otimes C) \otimes D$ and not from $(A \otimes B) \otimes C$ similarly to the definition of Monoid Object ?

I think I don't really understand what we want to state with that diagram.

  • 2
    $\begingroup$ The surprising coherence theorem shows that if this diagram commutes and a few other small diagrams commute (usually shown before the pentagonal diagram), then "all diagrams commute": essentially it means that if this diagram commutes then you can do whatever you like with $\otimes$ and associativity. $\endgroup$ Nov 18, 2017 at 16:42
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    $\begingroup$ There's no "diagram" to be made with only three terms; $(A\otimes B)\otimes C\simeq A\otimes (B\otimes C)$ is part of the basic data of a monoidal category. But merely having an isomorphism there is not enough to guarantee that the above diagram commutes, though we certainly want it to if we're trying to capture a notion of associativity. $\endgroup$ Nov 18, 2017 at 17:37

1 Answer 1


(Here is an nlab link talking about what I describe below)

It's too cumbersome to ask for all possible ways to interpret a product to be identical, which is why we merely require there to be coherence isomorphisms — a natural isomorphism between any two ways to multiply out a product. This even including inserting and removing copies of the empty product!

The unitors $A \otimes I \cong A$ and $I \otimes A \cong A$ and the associator $(A \otimes B) \otimes C \cong A \otimes (B \otimes C)$ are examples of this. But we also expect, for example, there to be a coherence isomorphism

$$ (A \otimes (B \otimes I)) \otimes (C \otimes I) \cong (A \otimes (B \otimes C)) \otimes (I \otimes I)$$

The important thing, however, is that we can (and should) demand that coherence be unique — that any way of combining these coherence isomorphisms gives the canonical one.

For example, I could build another isomorphism between the two products via:

$$ (A \otimes (B \otimes I)) \otimes (C \otimes I) \cong (A \otimes B) \otimes C \cong (A \otimes (B \otimes C)) \otimes (I \otimes I)$$

Composing these two isomorphisms is guaranteed to be the same isomorphism as the one in the previous equation. The same is true if I construct an isomorphism out of the associator and unitor natural isomorphisms.

The pentagon is the smallest example of how to use the associator isomorphism to construct two distinct paths between interpretations of the product; the pentagon identity is the one asserting that the two paths give the same isomorphism.

I think it's clear that the unitors and the associator are enough to produce a coherence identity between any two ways of multiplying out a product. It's maybe somewhat surprising, though, that all of the coherence identities they have to satisfy are generated by the pentagon and the triangle (which relates the unitors and the associator).


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