Michael Hong wrote:

> So for this cap, we have \\( \cap_X (x,x') = \hom(x',x) = \cup_X (x,x')\\) ?

Right.

I find the arbitrary conventions in this course endlessly confusing, first because there are \\(2^n\\) options some fairly large value of \\(n\\) (like 10), second because there aren't completely standard choices in every case, and third because using Fong and Spivak's textbook I don't feel completely free to make the choices _I_ think are best!

For example: I don't like string diagrams going left to right; I like them going top to bottom. So when I say the 'cap'

\[ \cap_X \colon \textbf{1} \nrightarrow X \times X^{\text{op}} \]

is drawn like this:



I'm already annoyed that the cap isn't being drawn as an actual cap \\(\cap\\). And I'm also annoyed by the fact that this convention doesn't turn into my usual convention just by rotating the picture 90 degrees... I could explain that, but never mind.

But I can live with all this.

On a separate note, I really want the cap to go like this:

\[ \cap_X \colon \textbf{1} \nrightarrow X \times X^{\text{op}} \qquad (GOOD) \]

and not this:

\[ \cap_X \colon \textbf{1} \nrightarrow X^{\text{op}} \times X \qquad (EVIL) \]

because in the category of finite-dimensional real vector spaces we have a similar such thing for any object \\(V\\):

\[ \cap_X \colon \mathbb{R} \to V \otimes V^\ast \]

where \\(V \otimes V^*\\) is the space of linear transformations of \\(V\\) (or crudely speaking, 'square matrices') and \\(\cap_X\\) sends the number 1 to the identity matrix. It's purely a convention that we think of linear transformation of \\(V\\) as elements of \|(V \otimes V^\ast\\) rather than \\(V^\ast \otimes V\\), but this conventions works well if we multiply a vector by a matrix on the _left_, as we usually do.

(Here I'm hinting at, but not really explaining, a deep connection between profunctors and linear algebra. We already seen it before: our formula for composing profunctors looks just like the formula for matrix multiplication! So, I want my conventions in these various subjects to match up nicely... and this choice of the cap makes it work.)

Given that we have

\[ \cap_X \colon \textbf{1} \nrightarrow X \times X^{\text{op}} \]

it's easy to see that we must have \\(\cap_X(x,x') = \texttt{true}\\) iff \\(x' \le x\\), because this relation must remain true if we make \\(x \in X\\) _bigger_, and also if we make \\(x' \in X^{\text{op}} \\) _bigger_, which is the same as making \\(x' \in X\\) _smaller_.

I could talk for hours about these interlocking conventions, since I've spent my life worrying about them.... but this is enough!