A category $\mathbb{C}$ comprises the following data: a set $C_0$, a set $C_1$, a map $s : C_0 \to C_1$, a map $t : C_0 \to C_1$, a map $\rho : C_0 \to C_1$, and a map $\tau : C_2 \to C_1$, where $C_2 = \{ (g, f) \in C_1 \times C_1 : s(g) = t(f) \}$, and all these are required to satisfy various axioms.
We require $s(\rho(x)) = t(\rho(x)) = x$: $\rho : C_0 \to C_1$ is the witness of the reflexivity of this graph; in other words, $\rho(x)$ is a distinguished edge from $x$ to $x$.
The picture to draw here is a triangle: we have three vertices $x$, $y$, and $z$, and three edges $f$, $g$, and $\tau(g, f)$, where $f$ goes from $x$ to $y$, $g$ goes from $y$ to $z$, and $\tau(g, f)$ goes from $x$ to $z$.
@Jordan Hey, just keep doing. That was quite right and good. Only the last part wasn't correct. Take a look at it now and tell me if you can find where was the mistake.
Of course, traditionally $C_0$ is called the set of "objects", $C_1$ is called the set of "morphisms", $s$ is normally called $\textrm{dom}$, $t$ is normally called $\textrm{codom}$, $\rho(x)$ is normally written $\textrm{id}_x$, and $\tau(g, f)$ is normally written $g \circ f$... but I find this tends to put certain, uh, preconceptions in people's minds.
Conversely, every directed graph $\Gamma$ generates a "free" category, whose "morphisms" are just finite head-to-tail strings of directed edges of $\Gamma$.
The "identity morphism" is the empty string (for each vertex separately).
But if you have a strongly connected directed graph (with a unique loop on each vertex), then that's already a category.
No, wait, sorry. Don't ignore my previous comment: $s$ and $t$ give us the start and end point (respectively) of an edge in $C_1$. So why are they maps $C_0 \to C_1$ instead of $C_1 \to C_0$?
@PeterTamaroff Then afterwards, you were not second choice for the single voter who put checkmath first (who didn'r provide any alternative vote, by the way), nor third choice for any of the two voters who voted 1-Bill, 2-checkmath (their third votes went to Eric and Benjamin).
This seems pretty simple to me but I can't get it.
$$\int \sin^2 x \cos^2 x dx$$
$$\int (1-\cos^2 x) \cos^2 x dx$$
I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an id...
This seems pretty simple to me but I can't get it.
$$\int \sin^2 x \cos^2 x dx$$
$$\int (1-\cos^2 x) \cos^2 x dx$$
I know there is a rule in my book (with little explanation) that tells me when I had an odd and an even degree on two trig functions I should split the odd and convert it to an id...
