Let $f:X\to Y$. Prove that
$$\text{f is injective}\iff \forall A \subseteq X, \quad f(X-A)\subseteq Y-f(A)$$
My try:
For $\leftarrow$.
Let $f(x_1)= f(x_2)$ and take for $A=\{x_1\}$ then if $x_1\neq x_2$ then $x_2\in X-A$ so $f(x_2)\subseteq Y-f(x_1)=Y-f(x_2)$ so $x_1=x_2$
For $\rightarrow$
If $y\in f(X-A)$ then $f^{-1}(y) \in X-A $
$y\in f(X) $ and $f^{-1}(y)\notin A$
Since $f(X)\subseteq Y$ then
$$y\in Y-f(A)$$
I'm not sure for $\leftarrow$, but for $\rightarrow$ I think I'm wrong because didn't use injective
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Martin Sleziak
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Related: http://math.stackexchange.com/questions/511662/proving-fc-setminus-fd-subseteq-fc-setminus-d-and-disproving-equality – Asaf Karagila Nov 26 '16 at 21:38
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It is better to use $\setminus$ for the relative complements, not to mix up with the Minkowski addition. – A.Γ. Nov 26 '16 at 21:43
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The claim: If $y\in f(X-A)$ then $f^{-1}(y) \in X-A$ is not true in general, since $f^{-1}(y)$ may contain many points. However, for an injective function $f^{-1}(y)$ is a singleton. – A.Γ. Nov 26 '16 at 21:57
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See also Does $f(X \setminus A)\subseteq Y\setminus f(A), \forall A\subseteq X$ imply $f$ is injective ? and If $f$ is 1-1, prove that $f(A\setminus B) = f(A)\setminus f(B)$ – Martin Sleziak Nov 27 '16 at 14:33