Proof regarding compositions and proving bijective and inverses.

Question

Suppose that $f : A → B$ and $g : B → A$. Suppose also that $g\circ f = i_A$ and $f \circ g = i_B$. Then $f$ and $g$ are bijective and $g =f^{-1}$.

I came across this proof while studying for my final; any help would be greatly appreciated.

Answer 1

To prove that the functions are bijective, we just need to show that they are injective and surjective. Then we can complete the second part of the problem and show that $g=f^{-1}$.

Let's prove that $f$ is injective (the proof that $g$ is injective should be clear from this proof). Then suppose that $x, y \in A$ such that $f(a)=f(b)$. Then, by assumption, we can see that

$$ f(a)=f(b) \implies (g \circ f)(a)=(g \circ f)(b) \implies a=b $$

So, we have that $f$ is injective. Now let's show surjectivity. Consider an element $b \in B$. We know that

$$ (f \circ g)(b) = b \implies f(g(b))=b $$

Therefore, we have that there exists a value in $A$, $g(b)$, which maps to $b$. Therefore, $f$ is surjective and then we can also conclude that $f$ is a bijection.

As I've mentioned before, the proof that $g$ is also a bijection should be fairly straight forward. From there, it's fairly clear that $f$ and $g$ are inverses of each other and so then we say that $g$ is the inverse, because inverses of bijections are unique. Thus, we've proved the statement.

Good luck on your final.