proof about injection I have to proof that the function $f : X \rightarrow Y$ is an injection if and only if $\forall T \subseteq X$, $f(X\setminus T) \subseteq Y \setminus f(T)$.
I'm having some difficulties. First (1) I proof that if $f$ is an injection then $\forall T \subseteq X$, $f(X\setminus T) \subseteq Y \setminus f(T)$, successively I'll prove the inverse implication (2).
(1):
I want to show that a generic element of $f(X\setminus T)$ belongs to $Y \setminus f(T)$ too, but I don't know how to continue, it is not apparent to me how to use the injection of $f$.
(2)
EDIT as suggested by Mark, i'll try the direction 2.
We know that $\forall x \in X\setminus T$ and $\forall t \in T$, we have $x\neq t $, because x belongs to X but it does not belong to T. Now , if I choose $y \in f(X \setminus T)$, there is $x \in X \setminus T$ : $ y = f(x)$, because y is an element of the image of $X\setminus T$ through $f$. But we know also that $y \in Y\setminus f(T)$, so $y \notin f(T)$, this implies that $\forall t \in T$, $y\neq f(t)$. This reduces to $f(t) \neq f(x) $. Is this proof valid ?
 A: It's straight from the definitions. Let $y\in f(X\setminus T)$. We need to show that $y\in Y$ and that $y\notin f(T)$. Obviously $y\in Y$ because $f$ maps elements of $X$ to $Y$. Now assume that $y\in f(T)$. Then there is an element $x\in T$ such that $f(x)=y$. On the other hand $y\in f(X\setminus T)$ so there is also an element $z\in X\setminus T$ such that $f(z)=y$. Hence $f(x)=f(z)$ when $x\ne z$ which is a contradiction to $f$ being injective. So $y\in Y\setminus f(T)$. Now can you prove the other direction? 
A: Partial answer:
2) $\forall T \subset X$,  $f(X$ \ $T) \subset Y$ \ $f(T)$ 
$\Rightarrow$ $f$ injective.
Let $x \not = t$;
With $T=${$t$}:
$x \in X$ \ {$t$} ; and 
$f(x)\in f(X$ \ {$t$}) $ \subset Y$ \ $f(${$t$}).
Then
$f(x) \not \in f(${$t$}$)$, i.e. $f(x)\not =f(t)$, injective.
A: So we want to show $f$ is injective iff
$$\forall T \subseteq X: f[X\setminus T]\subseteq Y \setminus f[T]\tag{*}$$
So let $f$ be injective, $T \subseteq X$ and let $y \in f[X\setminus T]$, i.e.
$y=f(x)$ with $x \notin T$. By definition of the function, $y \in Y$, but I claim that also $y \notin f[T]$, because otherwise $y=f(t)$ for some $t \in T$, and as $x \notin T$, we have that $x \neq t$ but also $y=f(x)=f(t)$ contradicting that $f$ is injective. So $y \in Y\setminus f[T]$, and $(\ast)$ has been shown.
Suppose that $(\ast)$ holds. Suppose $f$ were not injective, so that we have that $x_1 \neq x_2$ in $X$ with $y:= f(x_1) = f(x_2)$. Then $f[X\setminus \{x_1\}] = Y$ (as the only point we could miss is $y$ but this is also assumed by $x_2 \in X\setminus \{x_1\}$) but $Y\setminus \{y\}$ is a proper subset of $Y$, so that $(\ast)$ fails for $T=\{x_1\}$ (or $T=\{x_2\}$ too).
This contradiction shows that $f$ is injective.
