Invertible function without onto I have come across a situation in differential calculus where my mentor said that a function is invertible if it is one one because we consider range of function as domain of function's inverse. What does it mean?
 A: When we restrict the codomain to the range of the function, then of course, the term onto can be neglected, and only by checking the one one condition, we can say whether the given function is invertible or not. Onto is only needed to be considered, when the image set i.e. range is not equal to the codomain.
A: The function $$\exp : \mathbf{R} \to \mathbf{R}, \exp(x) = e^x$$ is one-to-one but not onto because $e^x$ is never negative for real values of $x$. But the function $$\exp^+ : \mathbf{R} \to \mathbf{R}^+, \exp^+(x) = e^x,$$ where $\mathbf{R}^+$ is the set of positive real numbers, is onto.
For all $x \in \mathbf{R}$,
$$ \exp(x) = e^x = \exp^+(x) $$
so $\exp$ and $\exp^+$ are essentially the same function except that they have different codomains. If we use $\mathbf{R}$ as the codomain, we can't get an inverse function $\mathbf{R} \to \mathbf{R}$. But if we use $\mathbf{R}^+$ as the codomain, then we have the inverse function $\log : \mathbf{R}^+ \to \mathbf{R}$.
A: Perhaps this will clarify the gist of the matter:
Remember that the domain of a function is the set of all possible inputs to that function, and the range is the set of all possible outputs of that function. A function is called one-to-one if for every output of the function, there is only one input to the function having that output (one input for every output, and one output for every input, hence the name one-to-one).
The inverse of a function maps the outputs of the function to its inputs. Thus it maps things in the range to things in the domain of the original function, hence the domain of the inverse is the range of the original and vice-versa. But remember that all functions (not just the one-to-one functions) must map one input to one output; a function cannot map one input to two or more outputs. Thus for the inverse of a function to itself be a function, the original function must be one-to-one because the outputs of the original function are the inputs of the inverse function. Being one-to-one ensures that there aren't two inputs to the original function mapping to the same output.
