Proving the injectivity of a function starts with lines similar to the following:
Assume that $f(x_{1}) = f(x_{2})$. If $x_{1} = x_{2}$, then $f$ is an injection.
Checking for the surjectivity of a function requires solving for the inverse and so on. Is there a similar way to prove the surjectivity of a function using a process similar to the one above?
Best Answer
Define $f : X \to Y$. Assume $y \in Y$. If you can show there exists at least one $x \in X$ such that $f(x) = y$, then you can show that $f$ is surjective.
Alternatively, say you define a function $g : Y \to X$. If you can show that $(f \circ g)(y) = y$ for all $y \in Y$, then $g$ is a right inverse to $f$, and thus surjective.
Alternatively, let $f^{-1}(B)$ denote the preimage of $B$, i.e. it is not an inverse, but rather
$$f^{-1}(B) = \{ x \in X \mid f(x) \in B \}$$
Then if for all $B \subseteq Y$ we have $f(f^{-1}(B)) =B$, then $f$ is surjective. Similarly, for all $B,C$ such that $B\subsetneq C \subseteq Y$, $f$ is surjective if $f^{-1}(B) \subsetneq f^{-1}(C)$.
All four of these are equivalent to surjectivity for a function $f$. Though, in my opinion, the first two are the main ones of utility.