1.3.10 Exercises

4 Let f :: A -> B be an isomorphism. Show that it's inverse is unique.

First, I will suppose that the inverse of f is not unique, that is there exists two unique arrows a,b :: B -> A such that:

- f . b = idA, b . f = idB
- f . a = idA, a . f = idB

Therefore:

- f . b = f . a
- a . (f . b) = a . (f . a)
- (a . f) . b = (a . f) . b
- idB . b = idB . a
- b = a

5 Show that if f' is the inverse of f :: A -> B and g' is the inverse of g :: B -> C, then (f' . g') is the inverse of (g . f).

(Notational note: In this exercise I'm using prime (') to denote inversion)

- (g . f)' . (g . f) = idA
- ((g . f)' . g) . f = idA
- (g . f)' . g = f'
- (g . f)' = f' . g' QED

I don't really understand this question ... Does the category 2 satisfy this? The only non-identity arrow is trivially both epic and monic and has no inverse. So, yeah.

That's it for this bit of questions! Hopefully I can keep this up for at least the next exercise set.

## No comments:

Post a Comment