I'm reading a book and it proves the following theorem:

Given a tensor category $(\mathcal{C}, \otimes, I, a, l,r)$, there exists a strict tensor category $\mathcal{C}^{str}$ (that is, the constraints $a,l,r$ are all identity transformations) such that $\mathcal{C}$ and $\mathcal{C}^{str}$ are tensor equivalent.

My book then claims:

This theorem implies Mac Lane's coherence theorem which states that in a tensor category any diagram built from the constraints a, l, r, and the identities by composing and tensoring, commutes.

How exactly does Mac Lane's coherence theorem follow from this?


Let $(\mathcal{C}, \otimes, I , a, l,r)$ be a tensor category. Consider a diagram in $\mathcal{C}$ built from the constraints $a,l,r$ and the identities by composing and tensoring.

Next, choose a strict tensor category $\mathcal{C}^{str}$ such that $\mathcal{C}$ is tensor equivalent with $\mathcal{C}$. Let $F$ be a tensor equivalence implementing the isomorphism $\mathcal{C} \stackrel{\cong}\to \mathcal{C}^{str}$. I can apply $F$ on my given diagram but I don't think it is as simple that $F$ preserves the relevant structure.

  • $\begingroup$ There are at least two ways to construct an equivalent strict tensor (or monoidal) category from $\mathcal{C}$: form finite sequences of objects of $\mathcal{C}$, or consider the category of "right $\mathcal{C}$-module endomorphisms of $\mathcal{C}$", i.e. endofunctors $F$ with a natural isomorphism $F(X) \otimes Y \cong F(X \otimes Y)$ that is compatible with the monoidal structure. $\endgroup$ – Geoffrey Trang Jun 19 at 0:11
  • $\begingroup$ @Geoffrey Trang. My book uses the finite sequences. But how is that relevant? $\endgroup$ – user745578 Jun 19 at 7:59

Technically this is only true for free diagrams, i.e., diagrams which take into account only the associator, right and left unitors, and identity morphisms, but not any other particular isomorphism (or equality between objects) that could be present in your particular category (I'll give a counterexample below).

The actual theorem is then that in the free tensor category generated by a set, every diagram commutes.

This is easy to see as a corollary of the strict tensor equivalence result: the free tensor category $FX$ is tensor equivalent to a strict one $F_sX$, which is discrete (composed only of identity transformations). Therefore:

  1. Any diagram of $FX$ is embedded as a diagram of $F_sX$.
  2. Any two parallel morphisms in $F_sX$ are equal.
  3. By the previous item, any diagram of $F_sX$ commutes.
  4. As a consequence of 1 and 3, any diagram of $FX$ commutes.

Now let us see a tensor category with a noncommutative diagram, an example due to Isbell:

Consider the skeleton category of the category Set (which is the full subcategory of cardinal numbers), made a tensor category with the cartesian product $\times$ as tensor product. Let $N$ denote the object with denumerable cardinal; then $N=N\times N$ and both projections $p_1,p_2:N\times N\rightarrow N$ are epi morphisms $N\rightarrow N$. We show that the associator isomorphism $a:(N\times N)\times N=N\times(N\times N)$ is not the identity. Due to the naturalness of $a$, if it was the identity, for any three $f,g,h:N\rightarrow N$ we would have $f\times(g\times h)=(f\times g)\times h$. But then $$fp_1=p_1(f\times(g\times h))=p_1((f\times g)\times h)=(f\times g)p_1:N\rightarrow N$$ implies that $f=f\times g$ for all $f,g$ (because $p_1$ is epi). Similarly $p_2$ being epi means that $g=f\times g=f$, which is a contradiction. So $a_{N,N,N}$ is not the identity. Thus the diagram Noncommutative tensor diagram

is not commutative. The problem here is that $N=N\times N=(N\times N)\times N$ is an identity which is not present in a free tensor category.

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