Inverses / Bijections

Question

Let $f:A\to B$, and $g:B\to A$ such that

$$ g(f(a))=a \ ,\ \forall\ a \in A, $$ and $$ f(g(b))=b\ ,\forall\ b \in B. $$ Does this mean that $f,g$ are inverses and bijections?

Bests

Answer 1

Yes. (I mean, what more can we really say?)

Hope that helps,

Answer 2

$f:A\to B, g:B \to A$

$afg=a$ then $g=f^{-1}$ and $a_1fg=a$,$a_2fg=a_2$ means that if $a_1fg=a_2fg\implies a_1=a_2$, hence $fg$ is injective and we know that means that $f$ is injective.

$(gf)$ is surjective, meaning, $\forall b\in B,\exists b\in B$ s.t. $bgf=b$

This is obviously true, and means that $f$ is surjective:

Proof: $\forall b\in B,\exists b\in B$ s.t. $bgf=b$

$(bg)f=b$ is surjective, hence $(bg)=(a\in A)$ and $af=b$

$(\forall b\in B)(\exists a\in A)|(af=b)$

Hence $f$ is surjective.

Since $f$ is injective and surjective, $f$ is a bijection.

If $f$ is bijective, we know it's inverse is bijective. (Which can be proved by repeating the above for $g$)

Hence these are bijective inverses.