(Abstract Algebra 1) Determining if a Function is Invertible

A function is invertible if and only if it is bijective.

[@iitjamganesha273]

Thank you Sir... you explained very well... I had been trying to understand that from book but was unable to comprehend ... I am from India thank you ...

[@lucjandespiegeleire454]

I think you forgot to mention in your proof when you suppose that f is bijectivef:A->B or is this optional?

[@monikhan1449]

lucjan de spiegeleire

[@Love_Hope_from_Above]

Prof. Learnifyable: The video "Determining if a Function is Invertible" is another wonderful lesson you have created in early February 2014. I really like the scratch of proof you did before writing a formal proof: You showed the thought process for creating a proof step-by-step. Great job! The natural next step after this lesson would be isomorphism since we require a bijective map from function f to function g to show an isomorphic structure. As I am hungry for more abstract-algebra lessons from you, can you kindly produce more abstract-algebra videos in April 2014? I am especially interested in isomorphic structures and generators of cyclic groups. Thank you very much -- you're a marvelous communicator of higher math (abstract algebra and beyond)! > Benny Lo Calif. 4-8-2014

[@learnifyable]

I'd like to make videos on both isomorphisms and cyclic groups. I would like to make a few videos on some number theory topics (gcd, lcm, etc.) before I tackle cyclic groups, however, since many of those concepts are used in the proofs. Don't worry--I haven't forgotten about abstract algebra. You will see some more abstract algebra videos soon!

[@ptyamin6976]

At 11:43, on the Surjective-side proof, you say: If we let a=f^-1(b), then we have found and element a e A such that f(a) = b How do you know that f^-1(b) is as element of A? Is it because you assume f is invertable? Would you not be able to make that step of f were not invertable?

[@thetheoreticalnerd7662]

I am a bit late, but he was proving that if f is invertible, then f is bijective. So it is assumed that f is invertible.

[@elelelel5756]

@@thetheoreticalnerd7662 a bit, lol

[@kqnrqdtqqtttel1778]

@@elelelel5756 XD

[@brendanmccann5695]

@@thetheoreticalnerd7662 I think I'm with PT Yamin on this. It seems to me that the proof relables f^-1(b) as "a". Doesn't this assume that f^-1(b) exists? Mustn't this be proven?

[@kqnrqdtqqtttel1778]

@@brendanmccann5695 Huh？The proof is an if and only if proof. First, you assume f is invertible and then show that it is bijective. Then, you assume that f is bijective and show that it is invertible.