To that end I'm working through Pierce's Basic Category Theory for Computer Scientists. So far, category theory isn't hard conceptually - but the concepts are so abstract that I find it difficult to internalize them. But then, it's not like I spent a lot of time in the math dept. back when I was in school (as much as I liked the mathematical tools I used).

Although I shouldn't be going on about how "not hard" it is while I'm still in chapter one.

Onward!

1.3.10 Exercises

2 Show that in any category, if two arrows f and g are both monic then their composition (g . f) is monic. Also, if (g . f) is monic then so is f.

The first part:

- Let (g . f) . a = (g. f) . b
- -> g . (f . a) = g . (f . b)
- Because g is monic -> f . a = f . b
- Because f is monic -> a = b

The second part:

Suppose that (g . f) is monic.

Let's further suppose that there exists two arrows a and b such that:

f . a = f . b, a \= b

- g . (f . a) = g . (f . b), a \=b, via arrow composition
- (g . f) . a = (g . f) . b, a\=b, via associativity of composition

3 Dualize the previous exercise: state and prove the analogous proposition for epics.

Along the same lines as the previous proof:

- Let (g . f) be epic.
- Let there exist two arrows a and b such that a . g = b . g and a \= b.

a . (g . f) = b . (g . f), a \= b via associativity of composition

My assumptions are in contradiction. Therefore, if (g . f) is epic, there exist no two arrows a and b such that a . g = b . g and a \=b - that is, g is epic.

More to come. Let me know if my math is crap.

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