I've a math problem that I can't solve which is:
How do I prove that if a function is bijective then it's inverse is also bijective?
I know I've to prove that if then So then the inverse is injective.
Then to prove a function is surjective I've to show that each element of is the image of at least one element of
If I can show these two to be valid then the inverse is bijective.
Can anyone kindly tell me how do I show this?
How do I show that a inverse of a bijective function is injective and surjective?
it's not obvious that the inverse of a bijective morphism is a bijective morphism. in fact, in the category of topological spaces (where "morphism" means homeomorphism) this is false. it just happens to be true in SOME categories (Set is one of them).
the definition only tells us a bijective function has an inverse function. it doesn't explicitly say this inverse is also bijective (although it turns out that it is). it's pretty obvious that in the case that the domain of a function is FINITE, f-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). with infinite sets, it's not so clear.
the problem is with "surjectiveness". it only asserts the existence of a pre-image for a co-domain element, it doesn't say how we should go about producing one. to even say we CAN always produce one is tantamount to the axiom of choice (something one "just has to take on faith"). this is in stark contrast to the situation with injective functions, where there is a clear-cut test: if f(x) = f(y), see if x = y is true.
insofar as "proving definitions go", i am sure you are well-aware that concepts which are logically equivalent (iff's) often come in quite different disguises. some texts define a bijection as a function for which there exists a two-sided inverse. some texts define a bijection as an injective surjection. the equivalence of the two isn't "something you can prove", in fact there are those who doubt it's even TRUE.
Thanks Deveno and Plato for answer.
Sorry if I'm missing something obvious Deveno. If is surjective why is it enough to only show injective or surjective in order to prove that the inverse function is bijective for finite set?
To prove bijection don't we need to prove that the function is both injective and surjective? Isn't that the definition of bijection?
Can you kindly shed light on this?