Answer to Question #174753 in Linear Algebra for Hetisani Sewela

Question

 Let f : A → B be a one-to-one correspondence.

1. Prove that f^-1 is a function.

2. Prove that f^-1 is one-to-one.

3. Prove that f^-1 is onto.

4. Conclude that f^-1 : B → A is a one-to-one correspondence.

Answer 1

1) By the theorem that states that a function is bijective if and only if it inverse exist, we can safely conclude that $f^{-1}$ exist and is indeed a function.

2) Let $x,y \in B$ . Consider

$f^{-1}(x) = f^{-1}(y)$

$f(f^{-1}(x)) = f(f^{-1}(y)) \\ f \circ f ^{-1}(x) = f \circ f ^{-1}(y)\\ x=y$

as desired.

3) Let $v \in A$

Then by surjectivity of f, $a = f(v)$ for some $a \in B$

So we have that $f^{-1}(a) = f^{-1}(f(v)) = v$

Hence $f^{-1}$ is surjective.

4) Since $f^{-1}$ is both injective and surjective, we can conclude it is a one-one correspondence.