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Friday, July 06, 2007

More from Pierce's Category Theory

Continuing the last set of exercises:

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
If f is an isomorphism, it's inverse is unique.

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
6 Find a category containing an arrow that is both a monomorphism and an epimorphism, but not an isomorphism.

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.

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