Category Theory 2.1: Functions, epimorphisms
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange askers link it when they get stuck on the motivation and proof behind the definition of an epimorphism, citing a specific moment near the end of the video.
Recommended by 3 people across 3 different months, 2017 to 2022.
What it is
Lecture 2.1 of Bartosz Milewski's category theory series, covering functions between sets and how surjections generalize to epimorphisms in an arbitrary category. It walks through the argument that a function is epic exactly when it is onto, using pairs of functions that agree on the image but differ outside it.
What people said
Recommends Isaac · 2017-05 · Mathematics Stack Exchange · on “Help understanding the concept of an epimorphism in this category video lecture series?”
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 [this video] 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: @ 42:5
Recommends user3578468 · 2017-07 · Mathematics Stack Exchange · on “Definition of Epimorphism in Category Theory”
Given a morphism $f : A \to B$ , $f$ is en epimorphism iff $\forall g_1, g_2 : B \to C. g_1 \circ f = g_2 \circ f\implies g_1 = g_2$ In words, a morphism $f : A \to B$ is epic if it has the property, that if any two morphisms $g_1, g_2 : B \to C$ are equal composed with $f$ then they are also equal on their own. I don't understand the point of this definition. In fact, I looked up the above definition from Wikipedia after also struggling with the definition given in this lecture: [LINK] $$\left[\forall c. \forall g_1, g_2 : b \to c. g_1 \circ f = g_2 \circ f \implies g_1 = g_2 \right]\implies f : a \to b \text{ is an epimorphism}$$ The motivation for this has to do with surjections and is gi
Recommends kingpin · 2022-10 · Mathematics Stack Exchange · on “Establishing the concept of epimorphism in category theory from surjective function in set theory”
I was reading here [LINK] to understand how to establish the concept of epimorphism in category theory from surjective function in set theory, got stuck here [this video] . Definition: A function $f:A\to B$ is an epimorphism if for all sets $Z$ and all functions $a_1,a_2:B\to Z$ , $a_1\circ f=a_2\circ f\implies a_1=a_2$ . If $f:A\to B$ is an epimorphism, then by the above definition we can choose two functions $a_1,a_2:B\to Z$ as described above such that $a_1\circ f=a_2\circ f\implies a_1=a_2$ . Now assume that $f$ is not surjective. Then there is some $b\in B$ such that $b \not\in\text{Im}f$ . If so, it's possible that $a_1(b)\ne a_2(b)$ so the implication $a_1=a_2$ does not hold. Therefore, fo