Is there a straightforward "mechanical method" to check whether a function is bijective or has an inverse? Is there a straightforward "mechanical method" to check whether a function is bijective or has an inverse? I have in mind here something like the truth table as a test for validity in logic.
Thanks in advance!
 A: Yes, the method is called verifying the axioms.
Given $f:X\to Y$ and suppose $y\in Y$ is some generic point, see if there is always an $x$ such that $f(x)=y$ to show it's surjective.
Then suppose that $f(x)=f(x')$ and see if it is absolutely certain that $x=x'$. This would show it is injective.
The truth or falsity of these statements is what needs to be checked.
A: A mechanical method is to consider the equations $y=f(x)$ and see how many solutions in $x$ they have for a fixed $y$.
We will say that a  function $f:X\longrightarrow Y$ has an inverse if there exists $g:f(X)\longrightarrow X$ such that $f\circ g=Id_{f(X)}$ and $g\circ f=Id_X$. This is equivalent to $f$ being injective, namely $x_1\neq x_2$ implies $f(x_1)\neq f(x_2)$. In other terms, for every $y\in Y$, the equation $y=f(x)$ has at most one solution in $X$.
If you have the graph of $f$, this is equivalent to the following geometric condition: all the level sets $\{x\in X\;;\; f(x)=y\}$ contain at most one point. In the real-real case, very concretely, it means that every horizontal line has at most one intersection point with the graph of $f$.
Now $f:X\longrightarrow Y$ is bijective if it has an inverse and $f(X)=Y$. This is equivalent to injective+onto. This means that for every $y\in Y$, there exists (by onto) exactly one (by injective) solution $x\in X$ to $y=f(x)$. It is denoted by $f^{-1}(y)$ and this defines the inverse function $f^{-1}:Y\longrightarrow X$. Geometrically, this means that the every level set contains exactly one point. In the real-real case, this allows you to take the symmetric of the graph of $f$ with respect to the diagonal $\{x=y\}$ and still obtain the graph of a function: that's the graph of $f^{-1}$.
