"Wrong" strictification of symmetric monoidal categories

Question

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

and

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

So what is going wrong?

Answer 1

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.

Answer 2

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 $\langle \textbf{Ab}, \otimes, \cdots \rangle$.