Michael wrote:

> If you strictify a monoidal category, is it somehow related to a skeletal category?

No. I'll start by refering you to [comment #12 on Lecture 72](https://forum.azimuthproject.org/discussion/comment/20709/#Comment_20709). Then I'll say some more...

> Ken wrote:

> > Theorem 1, MacLane's theorem - is this similar to how we got a "skeletal" poset out of a preorder by collapsing all the isomorphisms?

> **NO!!!**

> That's a mistake every newcomer makes. It's a natural guess, but fact Mac Lane's theorem is proved by making the monoidal category very 'fat', the opposite of skeletal.

> Unlike a preorder, a skeletal category still has lots of isomorphisms: it's just that they go from an object to itself. So, to make the associator and unitor isomorphisms between identity morphisms, it doesn't help to make the category skeletal. In fact it makes it harder!

> It would take quite a while to explain this, and I don't have the energy, but it's fairly rare for a monoidal category to be both strict and skeletal. To get a vague sense of how tricky things are: the category of finite sets with \\(\times\\) as its monoidal structure is monoidally equivalent to one that's both strict and skeletal, but not the category of all sets.

> If you strictify a monoidal category, is it somehow related to a skeletal category?

No. I'll start by refering you to [comment #12 on Lecture 72](https://forum.azimuthproject.org/discussion/comment/20709/#Comment_20709). Then I'll say some more...

> Ken wrote:

> > Theorem 1, MacLane's theorem - is this similar to how we got a "skeletal" poset out of a preorder by collapsing all the isomorphisms?

> **NO!!!**

> That's a mistake every newcomer makes. It's a natural guess, but fact Mac Lane's theorem is proved by making the monoidal category very 'fat', the opposite of skeletal.

> Unlike a preorder, a skeletal category still has lots of isomorphisms: it's just that they go from an object to itself. So, to make the associator and unitor isomorphisms between identity morphisms, it doesn't help to make the category skeletal. In fact it makes it harder!

> It would take quite a while to explain this, and I don't have the energy, but it's fairly rare for a monoidal category to be both strict and skeletal. To get a vague sense of how tricky things are: the category of finite sets with \\(\times\\) as its monoidal structure is monoidally equivalent to one that's both strict and skeletal, but not the category of all sets.