Michael wrote:

> It seems like there should be a skeletal category lurking in there since we are collapsing all isomorphic objects into one object.

No, we're not. The actual proof of Mac Lane's strictification theorem proceeds by creating tons _more_ isomorphic objects. You start with a monoidal category \$$\mathcal{C}\$$. Then you create a new monoidal category \$$\mathrm{str}(\mathcal{C})\$$ whose objects are _lists_ of objects in \$$\mathcal{C}\$$. The tensor product of two lists

$$(c_1, \dots, c_m)$$

and

$$(d_1, \dots, d_n)$$

is the list

$$(c_1, \dots, c_m, d_1, \dots, d_n )$$

It's easy to see that this tensor product is strictly associative. A morphism in \$$\mathrm{str}(\mathcal{C})\$$ from

$$(c_1, \dots, c_m)$$

and

$$(d_1, \dots, d_n)$$

is defined to be a morphism

$$f \colon c_1 \otimes (c_2 \otimes (c_3 \otimes \cdots )) \to d_1 \otimes (d_2 \otimes (d_3 \otimes \cdots ))$$

in \$$\mathcal{C}\$$.

It takes some work to make \$$\mathcal{C}\$$ into a monoidal category and show that \$$\mathrm{str}(\mathcal{C})\$$ is monoidally equivalent to \$$\mathcal{C})\$$.

But my point here is \$$\mathrm{str}(\mathcal{C}\$$ is not skeletal! In this category

$$(c_1, \dots, c_m)$$

and

$$(d_1, \dots, d_n)$$

are isomorphic iff

$$c_1 \otimes (c_2 \otimes (c_3 \otimes \cdots ))$$

and

$$d_1 \otimes (d_2 \otimes (d_3 \otimes \cdots ))$$

are isomorphic in \$$\mathcal{C}\$$. That happens a whole lot! For example, the two-element list

$$(c_1,c_2)$$

is isomorphic to the one-element list

$$(c_1 \otimes c_2) .$$