# Bijection, Surjection, defining inverse, proof

Prove both the following:

1. If f : A → B is bijective, an inverse g of f exists. (You will have to define such a g, show it is a well-defined function, and show that it really is the inverse of f.)

2. Show that if f : A → B has an inverse g, then f is bijective.

(1) Definition

Let $$f: A\to B$$. We say that $$f$$ is surjective if for all b ∈ B, there exists an a ∈ A such that f(a) = b.

We say that f is injective if whenever f(a1) = f(a2) for some a1, a2 ∈ A, then a1 = a2.

We say that f is bijective if it is both injective and surjective.

For the first proof:

1. let $$f: A\to B$$ be bijective.
2. let $$g: B\to A$$ where y ∈ B

Since f is surjective, there exists an [x] ∈ A such that f(x) = y

1. let $$g(y) = x$$

Because f is injective, x is unique thus, g is well-defined.

would this be valid for the first part?

• The two questions are just one if and only if statements. For $(1)$, after defining $g$, note that $f(a)=b\implies g(b)=a$ and since $f$ is injective, then $a$ is unique and thus $g$ is bijective. It remains to show that $g$ is the inverse of $f$ and is well defined. Commented Jan 21, 2021 at 6:34