- Let $X$, $Y$ be sets and let $f:X \to Y$ be a map. For any subset $C \subseteq Y$ of the codomain $Y$, we consider its preimage $f^{−1}(C) \subseteq X$ in the domain $X$: $f^{−1}(C) := \{x \in X : f(x) \in C\}$.
(Note: the map $f$ here is not assumed to be invertible; thus the notation $f^{−1}$ here denotes “preimage”, not the inverse of $f$ .)
Show that $f$ is injective if and only if for any subset $C \subseteq Y$ such that $C$ is empty or a singleton, its preimage $f^{−1}(C) \subseteq X$ is empty or a singleton.
so if $C$ is empty or a singleton, the element will only hit at most one target on the range. because whatever we ask of the empty set will be true, so what are the steps i need to show in order to prove that it is injective. (injectivity, F(x1) = F(x2) in Y. right?
Hint:
To prove injectivity of $f$ it is enough to prove that $f(x_1)=f(x_2)$ implies $x_1=x_2$.
If $f(x_1)=y=f(x_2)$ have a look at $f^{-1}(\{y\})$ wich is the preimage of a singleton.
edit (on request):
Let it be that $f^{-1}(C)$ is empty or is a singleton whenever $C\subseteq Y$ is a singleton. To be shown is that $f$ is injective. Let it be that $x_1,x_2\in X$ with $f(x_1)=f(x_2)$. Set $C:=\{f(x_1)\}$ is a singleton so we are allowed to conclude that $f^{-1}(C)$ is empty or a singleton. The elements $x_1,x_2$ both belong to this set so it is not empty. Then it is a singleton and $f^{-1}(C)=\{x_1\}=\{x_2\}$ so that $x_1=x_2$. Proved is now that $f$ is injective.