Is the inverse of a function bijective as well? Why?

Question

I know that if the function is bijective we can define the inverse function. But how do we know that this inverse function is bijective as well? Should we proof it?

Answer 1

This is one of those "the proof is too small to fit in the margin" type propositions.

f:A->B being bijective means if $f(x) = f(y)$ then $x = y$ and f(A) = B.

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

So $f^{-1}$ is injective.

If $x \in A$ then $f(x) \in f(A) = B$ then $x \in f^{-1}(B)$. If $y = f^{-1}(B)$ then there exists a $f(y) \in B$ so that $f^{-1}(f(y)) = y \in A$. So $f^{-1}(B) = A$.

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

Answer 2

A quick google search leads me to the proof wiki which provides a proof. Basically the inverse of the inverse of a function is the function itself. This tells us that the inverse function has a inverse and is therefore bijective.

Answer 3

Let $f: A \to B$ be bijective. Then there exists a function $g: B \to A$ satisfying $$ (g \circ f)(x) = x $$ for all $x \in A$, and $$ (f \circ g)(y) = y $$ for all $y \in B$. To see that $g$ is bijective, we show that $g$ is injective and surjective.

Injective. Consider two $y_1, y_2 \in B$, and suppose that $$ g(y_1) = g(y_2). $$ Since $g(y_1), g(y_2) \in A$, we can fix $x_1, x_2 \in A$ such that $$ g(y_1) = x_1, \qquad g(y_2) = x_2, $$ with the assumption that $x_1 = x_2$. But since $f$ is a well-defined function, it follows that $$ f(x_1) = f(x_2). $$ Substituting $g(y_1)=x_1$ and $g(y_2) = x_2$, we see that $$ (f \circ g)(y_1) = y_1 = y_2 = (f \circ g)(y_2), $$ allowing us to conclude that $g$ is injective.

Surjective. Let $x \in A$ be arbitrary. Then $f(x) = y$ for some $y \in B$. Since $g$ is well-defined, $g(y) = z$ for some $z \in A$. Specifically, we have that $$ z = g(y) = (g \circ f)(x) = x, $$ from whence surjectivity follows.

We conclude that $g$ is bijective. Such a function is commonly denoted $f^{-1}$.

Note: We didn't use the fact that $f$ is bijective anywhere in this proof. Why not? It turns out that the assumption that $g$ exists at all is equivalent to the assumption that $f$ is bijective.