[Last time](https://forum.azimuthproject.org/discussion/2334/lecture-72-monoidal-categories#latest) I explained monoidal categories, which are a framework for studying processes that we can compose and tensor. We can do a lot with monoidal categories! For example, if we have a monoidal category with morphisms

$\Phi \colon a \to c \otimes d$
$\Psi \colon d \otimes b \to e \otimes f$
$\Theta \colon c \otimes e \to g$

then by a combination of composing and tensoring we can cook up a morphism like this: which goes from \$$a \otimes b\$$ to \$$g \otimes f\$$. This sort of picture is called a **string diagram**, and we've seen plenty of them already.

We don't _need_ to use string diagrams to work with monoidal categories:

**Puzzle 281.** Describe the morphism in the above string diagram using a more traditional formula involving composition \$$\circ\$$, tensoring \$$\otimes\$$, the associator \$$\alpha\$$, and the left and right unitors \$$\lambda\$$ and \$$\rho\$$.

However, they make it a lot easier and more intuitive!

An interesting feature of string diagrams is that they hide the the associator and the left and right unitors. You can't easily see them in these diagrams! However, when you turn a string diagram into a more traditional formula as in Puzzle 281, you'll see that you need to include associators and unitors to get a formula that makes sense.

This may seems strange: if we need the associators and unitors in our formulas, why don't we need them in our diagrams?

The ultimate answer is 'Mac Lane's strictification theorem'. This says that every monoidal category is equivalent to a one where the associator and unitors are _identity_ morphisms. So, we can take any monoidal category and replace it by an equivalent one where the tensor product is 'strictly' associative, not just up to isomorphism:

$(x \otimes y) \otimes z = x \otimes (y \otimes z)$

and similarly, the left and right unit laws hold strictly:

$I \otimes x = x = x \otimes I$

This lets us stop worrying about associators and unitors. String diagrams are secretly doing this for us!

Often people use Mac Lane's strictification theorem in a loose way, simply using it as an excuse to act like monoidal categories are all strict. That's actually not so bad, if you're not too obsessed with precisoin.

To state Mac Lane's strictification theorem precisely, we first need to say exactly what it means for two monoidal categories to be 'equivalent'. For this we need to define a 'monoidal equivalence' between monoidal categories. Then, we define a **strict** monoidal category to be one where the associator and unitors are identity morphisms. Mac Lane's theorem then says that every monoidal category is monoidally equivalent to a strict one.

If you're curious about the details, try my notes:

* [Some definitions everyone should know](http://math.ucr.edu/home/baez/qg-winter2001/definitions.pdf).

All the necessary terms are defined, leading up to a precise statement of Mac Lane's strictification theorem at the very end. But this theorem takes quite a lot of work to prove, and I don't do that! You can see a sketch of the proof here:

* John Armstrong, [The "strictification" theorem](https://unapologetic.wordpress.com/2007/07/01/the-strictification-theorem/).

But there's more! If all we have is a monoidal category, the strings in our diagrams aren't allowed to cross. But last time I mentioned [symmetric monoidal categories](https://en.wikipedia.org/wiki/Symmetric_monoidal_category), where we have a natural isomorphism called the **symmetry**

$\sigma_{x,y} \colon x \otimes y \to y \otimes x$

that allows us to switch objects, obeying various rules. This lets us make sense of string diagrams where wires cross, like this: **Puzzle 282.** Describe the morphism in the above string diagram with a formula involving composition \$$\circ\$$, tensoring \$$\otimes\$$, the associator \$$\alpha\$$, the left and right unitors \$$\lambda,\rho\$$, and the symmetry \$$\sigma\$$.

There is a version of Mac Lane's strictification theorem for symmetric monoidal categories, too! You can find it stated in my notes. This lets us replace any symmetric monoidal category by a **strict** one, where the associator and unitors _but not the symmetry_ are identity morphisms.

We really need the symmetry: it cannot in general be swept under the rug. That should be sort of obvious: for example, switching two numbers in an ordered pair really _does_ something, we can't just say it's the identity.

Again, please ask questions! I'm sketching some ideas that would take considerably longer to explain in full detail.

**[To read other lectures go here.](http://www.azimuthproject.org/azimuth/show/Applied+Category+Theory#Chapter_4)**