It is well-known that any symmetric monoidal category is equivalent to a strict symmetric monoidal category. The construction of this strict monoidal category is rather technical and it appears to me that there is a significantly easier way of obtaining an even better result. Of course this means almost certainly that I am making a mistake.

Given a symmetric monoidal category $S$, consider the category of isomorphism classes $T$. Explicitly, let $T$ be the full subcategory of $S$ which has one distinct object for every isomorphism class of $S$. The inclusion functor $F:T\to S$ is fully faithful and essentially surjective. Hence it is an equivalence of categories with some inverse $G$.

Now $T$ is strict symmetric monidal with operation $\oplus_T$ defined via

$$A\oplus_T B:=G(F(A)\oplus_SF(B))$$

This is

a) very simple


b) also commutativity is strict which appears to be better than the classical result.

So what is going wrong?

  • 4
    $\begingroup$ Why do you think the skeleton is strict? (Check the definition.) $\endgroup$
    – anon
    Apr 24, 2013 at 13:08

2 Answers 2


There are a couple of issues:
The first thing is that isomorphism classes don't form a category.

You may force them into being a category by picking a representative in each isomophrpism class, and then looking at the full subcategory that they span. But this construction depends on the choice of representative:

If you change your mind, and pick another set of representative, then there is no canonical equivalence of categories between those two full subcategories (there are many equivalences of categories between those two categories, but there is no canonical one).

Ok, maybe that's not too bad after all... This thing that you call "the category of equivalence classes" actually has a name: it's called a skeleton of the category.

Ok, now let's look at another, more serious isue:

A monoidal category (C,⊗) isn't just a category with a ⊗, it's also equipped with an associator α:(X⊗Y)⊗Z→X⊗(Y⊗Z).

If you restrict to a skeleton T⊂C of the category, then in order to equip that skeleton with a monoidal structure, you should pick for every object X of C an isomorphism $$ \beta_X:X\to \underline{X}, $$ where $\underline{X}$ is the chosen representative on the equivalence class $[X]$.

The monoidal product on $T$ is given by $$ X \underline\otimes Y := \underline {X \otimes Y} $$ and it has the following associator:

$$ (X \underline\otimes Y) \underline\otimes Z = \underline{X \otimes Y \otimes Z} = \underline{\underline{X \otimes Y} \otimes Z} $$

$$\xrightarrow{\beta^{-1} _ {\underline{X \otimes Y}\otimes Z}}\underline{X \otimes Y} \otimes Z \xrightarrow{\beta^{-1}_{\underline{X \otimes Y}}\otimes 1}(X \otimes Y) \otimes Z $$

$$\xrightarrow{\alpha} X\otimes (Y \otimes Z) \xrightarrow{1\otimes \beta_{Y \otimes Z}} X \otimes \underline{Y \otimes Z} $$

$$\xrightarrow{\beta_{X \otimes \underline{Y \otimes Z}}} \underline{X \otimes \underline{Y \otimes Z}} = \underline{X \otimes Y \otimes Z} = X \underline\otimes (Y \underline\otimes Z) $$

This associator goes from an object to itself, but is probably not trivial. So the resulting monoidal category is maybe skeletal, but certainly not strict.

  • $\begingroup$ Thanks. I wrote a few lines in an edit to indicate where exactly my mistake was. I was aware that the associators are different, but I for some reason assumed that the functor would allow for that. $\endgroup$ Apr 24, 2013 at 14:11
  • $\begingroup$ @SimonMarkett: The edit was incorrect, so I’ve removed it from there and will respond to it here. You had written “Edit: This is basically what André is saying in his answer. The skeletal category is certainly strict with a different associator. For some reason I thought that one had an additional natural transformation somewhere so that the associators don't have to be compatible in a strong sense. So we get an equivalence from a strict monoidal category to a given category in this way, but the functor is then not monoidal.” [cont’d] $\endgroup$ Aug 7, 2021 at 17:14
  • $\begingroup$ [cont’d] But the issue isn’t just the comparison functor — the skeletal category is not a strict monoidal category, since trying to use the identity $\mathrm{id}_{X,Y,Z}$ as an associator, one finds (as Zhen Lin’s answer shows) that it is not generally a natural transformation. $\endgroup$ Aug 7, 2021 at 17:17

There is a counterexample due to Isbell explaining why your construction does not always produce a strict monoidal category; see the closing remarks in [Categories for the working mathematician, Ch. VII, § 1]:

One might be tempted to avoid all this fuss with $\alpha$, $\lambda$, and $\rho$ by simply identifying all isomorphic objects in $B$. This will not do, by the following argument due to Isbell. Let $\textbf{Set}_0$ be the skeleton of the category of sets; it has a product $X \times Y$ with projections $p_1$ and $p_2$ as usual. If $D$ is a (the) denumerable set, then $D = D \times D$, and both projections of this product are epis $p_1, p_2 : D \to D$. Now suppose that the isomorphism $\alpha : X \times (Y \times Z) \to (X \times Y) \times Z$, defined as usual to commute with the three projections, were always the identity; it is then the identity for $X = Y = Z = D$; since $\alpha$ is natural, $f \times (g \times h) = (f \times g) \times h$ for any three $f, g, h : D \to D$. But $\times$ on functions is defined in terms of the projections $p_1$ and $p_2$ above, so $$f p_1 = p_1 (f \times (g \times h)) = p_1 ((f \times g) \times h) = (f \times g) p_1$$ and $p_1$ is epi, so $f = f \times g$. The corresponding argument with $p_2$ gives $f \times g = g$, hence $f = g$ for any $f, g : d \to D$, an absurdity. A similar argument applies to the skeleton of $\langle \textbf{Ab}, \otimes, \cdots \rangle$.

  • $\begingroup$ It certainly makes me feel better that I 'only' walked into a trap somebody else already 'set up'. +1 $\endgroup$ Apr 24, 2013 at 16:28
  • 2
    $\begingroup$ In the case of cartesian products, although we can't make the product strict, we get by because the functions to products are always defined by the universal property, and from products by the exponential law. The situation seems analogous for many tensor products because the tensor products are defined by a "bi-" property, like bilinear; similarly for chain complexes, cubical sets, and many others. So we deal without fuss with multilinearity and the associativity of such tensor products. These tensor products seem to be as coherently associative as the usual product. Is this naive? $\endgroup$ Apr 24, 2013 at 21:34
  • $\begingroup$ Though it's a decade later, I just thought I might note here that the universal property of many familiar monoidal products Ronnie mentions is nicely formalized in a multicategory, and then one does indeed get all the coherence "for free" out of the universal property. $\endgroup$ Sep 7, 2023 at 5:34

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