I have this problem on my hands right now:
Let $$f:A\to B$$ and $$g:B\to A$$ be two functions. If $$g \circ f(a) = a, \forall a\in A$$ and if $$f \circ g(b) = b, \forall b\in B$$ then $f$ is a bijection.
Here are the things that are needed for $f$ to be a bijection:
It has to be injective meaning: $$F(x)=F(y)\implies x=y$$ and it has to be surjective meaning: $$\forall b\in B, \exists a\in A : F(a)=b$$
If this is true, could anyone give me a head start on how to approach this with Induction? Or how to set this up. I have searched on the internet, but have not found this specific problem and I have no idea how to approach this.. Thank you.
So... you have the definition. So do it.
we have to prove:
1) $f$ is an injection.
we must prove if $f(x) =f(y)$ then $x = y$..
If $f(x) = f(y)$ then $g\circ f(x) = g\circ f(y)$. But we are told that $g\circ f(x) = x$ and $g\circ f(y) = y$. So $x=g\circ f(x) = g\circ f(y) = y$.
2) $f$ is a surjection.
we must prove if $y \in B$ there is an $x \in A$ so that $f(x) = y$.
We are told that $f\circ g(y) = y$ for all $y \in B$ so just let $x =g(y) \in A$. Then $f(x) = f\circ g(y) = y$.