Is the inverse of a function bijective as well? Why? 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?
 A: 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. 
A: 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.
A: 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.
