Preliminaries:
(1) I know that a mapping is bijective if and only if it is injective and surjective. (2) I also know that a mapping $f:X\rightarrow Y$ is surjective if and only if $f(X)=Y$, where $f(X)$ denotes the image of $X$ under $f$. (3) I've defined two sets $X$ and $Y$ to be equinumerous, written $X\sim Y$, iff there is a bijective mapping $f:X\rightarrow Y$.
Question:
In my notes I have the following statement (I can't remember where it's coming from, it's been a while…).
Let $A$ and $B$ be sets. A Mapping $f:A\rightarrow B$ is injective if and only if for all subsets $X\subseteq A$ we have $X\sim f(X)$, i.e., if for all $X\subseteq A$ there is a bijective map $g:X\rightarrow f(X)$, where $f(X)$ denotes the image of $X$ under $f$.
I've tried to come up with a proof for this statement but so far I didn't really succeed and so I've started to wonder if that theorem is actually true or not. I think I can proof the implication "$f$ is injective $\Longrightarrow$ $X\sim f(X)$", though.
Proof
$(\Longrightarrow)$ Let $f:A\rightarrow B$ be injective. Then because of (2) $\tilde{f}:X\rightarrow f(X)$ is surjective and therefore, because of (1), bijective. Then, from definition (3), it follows that $X\sim f(X)$.
I'm not quite sure about the first step in my proof though. When $f$ is injective, why is a map $\tilde{f}$ injective too? Furthermore, what about the $(\Leftarrow)$ direction of the proof? Anyone got an idea or should I rewrite the Theorem and make it an implication instead of an equivalence?
Best Answer
The equivalence is true and your proof of $\Rightarrow$ is correct.
When $f:A\to B$ is injective, so is its restriction $g:X\to B$ (if $f(x)=f(y)\Rightarrow x=y$ holds for all $x,y\in A$, it holds in particular for all $x,y\in X$), and corestricting $g$ to $f(X)$ doesn't affect this injectivity.
Proof of $\Leftarrow$: For any two distinct elements $x,y$ in $A$, apply the hypothesis "forall subset $X\subseteq A$ we have $X\sim f(X)$" to $X=\{x,y\}$. You get that $f(X)$ has exactly two elements, i.e. $f(x)\ne f(y)$. This proves injectivity.