Help understanding the concept of an epimorphism in this category video lecture series?

Question

I'm trying to learn category theory through self study, reading books and watching videos but I'm stuck on something that is perhaps stupid. Idk, but I've been thinking about it for days and I really don't want to move on unless I understand it.

So here's the link to the video where he begins this discussion. Here's what I understand:

He begins by saying, let's consider the case where $f:A \to B$ is not surjective, then there are elements in B that are not in the image of $f$. Check. Then he says let's consider two other functions, $g_1: B \to C$ and $g_2: B \to C$ which ONLY differ for elements OUTSIDE the image of $f$, thus $g_1 \circ f = g_2 \circ f$. Check.

Where I'm lost: @ 42:57 he says "but for the converse of this $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$"

Questions:

1) When he says $\forall g_1, g_2$, does he mean for every $g_1$ and $g_2$ that could possibly exist or just the ones that are in the category? Because if it's just the ones in the category, then what if the category only has the sets $A,B,$ and $C$ and the arrows $1_A,1_B,1_C, A \xrightarrow{f}B,B \xrightarrow{g_1}C, B \xrightarrow{g_2}C, g_1 \circ f, \text{and } g_2 \circ f $ and we know $f$ is not surjective and that $g_1$ differs from $g_2$ only outside the image of $f$, then it seems like $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$ is false because $g_1 \circ f = g_2 \circ f$ is true but $g_1 \neq g_2$. I'm super confused!

2) What is the proposition he is taking the converse of AND why is he taking the converse of this proposition if $Q \implies P$ does not always have the same truth value of $P \implies Q$?

3) I can't visualize $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$ being true would mean that $f$ is surjective. How can I see this?

Answer 1

1) any composable morphisms in the category. There is no confusion since the composition $g_1\circ f = g_1\circ f$ has to exists for the test.

2) he means the negation of (the negation of $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$). That is the negation of $\exists g_1,g_2 \; (g_1\circ f = g_2\circ f \nRightarrow g_1=g_2)$.

3) You can visualize the converse as in 2) which he does in 40:00+

Answer 2

I believe your confusion is due to the fact that at one point he is speaking of $g_1$ and $g_2$ being different, and then he speaks of a another condition in which they are the same. When he is speaking of there being a halo around $f$, he repeats that $g_1$ and $g_2$ are different:

42:28 "... but if they only differ outside, in this halo, the composition with $f$ will be the same."

42:52 "... even though $g_1$ is different from $g_2$."

Then he switches gears and says at 42:55, "So the converse of this...", and then he introduces the expression $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$. What is confusing about that is that diagram that he drew is no longer applicable.

So, to answer your questions:

1. Although the term surjective is usually associated to the category Set, the quantification $\forall g_1, g_2$ applies to any category, for all maps $f:A \to B$, $g_1: B \to C$ and $g_2: B \to C$.

What you said is true: if "we know $f$ is not surjective and that $g_1$ differs from $g_2$ only outside the image of $f$, then it seems like $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$ is false."

He never claimed otherwise. He is saying that if the expression is true, then it's surjective.

2. First he spoke of the existence of two unequal maps, $g_1$ and $g_2$, which preserve the image of $f$, that is, $g_1 \circ f = g_2 \circ f$. Then he switches gears and speaks of the converse of that (although he meant the negation):

$$\neg \exists g_1, g_2[g_1 \neq g_2 \land g_1 \circ f = g_2 \circ f]$$ $$\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$$

3. By definition, if $f$ were not surjective, there would exist $y \in B$ for which there would be no member $x \in A$ such that $f(x)=y$. That would be the case in where there would be a "halo", as he calls it.

When $\forall g_1, g_2 (g_1 \circ f = g_2 \circ f \implies g_1 = g_2)$ is true, there is no halo because the image of $f$ cannot imply anything about all possible maps $g_1$ and $g_2$ unless there is nothing else but the image in question. Otherwise, the halo would impose an unpredictable element which would make the implication impossible. That is precisely what he showed previously when he was talking about $g_1$ and $g_2$ being able to preserve the image of $f$ even though they are unequal outside of the image of $f.$