Inverse of a Function exists iff Function is bijective How to mathematically prove that inverse of a function, let's say, $f^{-1}$, exists, if and only if $f$ is bijective?
I know how to prove it using diagrams but I'm looking for a rather mathematical proof with arguments. Anybody who can help? Thanks.
 A: Here's a crazy answer, in the sense that it's very formal. First, for a pair of sets $A$ and $B$, define $\pi_1: A \times B \to A: (a, b) \mapsto A$, and similarly define the map $\pi_2$ to $B$. 
Now a function is (at least according to one formal definition) a triple 
$$
f = (X, Y, R)
$$
where $R$ is a subset of $X \times Y$ with two properties: 


*

*$\pi_1 ( R) = X$, i.e., every element of $X$ appears as the first element of some ordered pair in $R$, and 

*$\pi_1(x, y) = \pi_1(x, y') \Rightarrow y = y'$, i.e., each element of $X$ corresponds to at most one element of $Y$. 
In this context, the definition of "surjective" is $\pi_2(R) = Y$, and the definition of "injective" is $\pi_2(x, y) = \pi_2(x', y) \Rightarrow x = x'$. 
If we have a surjective and injective function, $f = (X, Y, R)$ we can build a new function $g = (Y, X, R')$, where $R' = \{ (y, x) | (x, y) \in R \}$. 
Clearly the properties of "surjective" and "injective" for $f$ turn into properties $1$ and $2$ for $g$; also pretty evident is that $g$ is the thing we usually call $f^{-1}$. 
A: Consider a function $f:X\to Y$. If its is not surjective, there exists $y\in Y$ without preimage. If it is not injective, there exists $y\in Y$ such that the preimage $f^{-1}(y)$ contains at least 2 elements...
A: Method:1
When $f:A\to B$ is not one-one,
$$
f(x_1)=f(x_2)=y' \text{ for } x_1\neq x_2\in A,y'\in B\\
\implies f^{-1}(y')=f^{-1}(f(x_1))=x_1 {\quad\&\quad} f^{-1}(y')=f^{-1}(f(x_2))=x_2
$$
For the inverse function $f:B\to A$, we have $f^{-1}(y')=x_1$ and $f^{-1}(y')=x_2$. Hence, $f^{-1}$ is not defined as we have two images for $x_1$ and $x_2$ for one preimage $y'$.
$\implies$ for $f$ to be invertible it must be one-one.
When $f:A\to B$ is not onto,
there exists atleast one $y'\in B$ which does not have a preimage in set $A$. When we define $f^{-1}:B\to A$, $y'$ becomes preimage which will not have any image in set $A$, thus $f^{-1}$ is not defined.
$\implies$ for $f$ to be invertible it must be onto.
Therefore,
$f$ is invertible iff $f$ is both one-one and onto.
Method:2
If $f:A\to B$ is invertible, then $f^{-1}\circ f=I_A$ and $f\circ f^{-1}=I_B$.
$$
f^{-1}\circ f(x_1)=f^{-1}\circ f(x_2)\implies f^{-1}(f(x_1))=f^{-1}(f(x_2))\\\implies x_1=x_2\text{ $\bigg($since $f^{-1}\circ f(x)=x, \forall x\in A\bigg)$}
$$
Thus, $f^{-1}\circ f$ is one-one $\implies$ $f$ is one-one.
$f\circ f^{-1}:B\to B$ defined as $f\circ f^{-1}(y)=y,\forall y\in B$. For all $y$ in the codomain $B$ have one preimage $y$ in the domain $B$.ie, Codomain of $f\circ f^{-1}$=Range of $f\circ f^{-1}=B$
Thus, $f\circ f^{-1}$ is onto $\implies$ $f$ is onto.
$\implies$ $f$ is bijective
A: Try using the left and right inverse concept:
Consider two functions: $f : A \to B$ and $g : B \to A$. When $f \circ g$ is the identity function on $A$, we say $f$ is a left inverse of $g$ and $g$ is a right inverse of $f$.
Then to see that a bijection has an inverse function, it is sufficient to show the following:


*

*An injective function has a left inverse.

*A surjective function has a right inverse.

*If a function has a left and right inverse they are the same function.

