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Prove that if a function f:X-->Y is injective, then [itex]f(X\setminus A) \subset Y\setminus f(A), \forall A \subset X[/itex].

This is how I did it:

If x_1 is in A, then y_1=f(x_1) is in f(A). Because the function is injective, we can pick (cut Y into pieces) f(A) and f(X\A) so, that their intersection is empty (adjust A if needed). So, if x_2 is not in A = if x_2 is in X\A, then y_2=f(x_2) in f(X\A). From this we get that y_2 is in Y\f(A), since Y\f(A)=f(X\A). So, I've shown using the injective property, that y_2 in Y\f(A) follows from y_2 in f(X\A).

Right? Wrong?

- Kamataat