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    injective and surjective

    For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.
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    Quote Originally Posted by rezona View Post
    For every function f, subset A of the domain and subset B of the codomain prove that If f is injective we have A = f −1(fA) and if f is surjective we have f(f −1B) = B.
    This is unreadable.
    It should read for the first one
    f\text{ is injective }\Longleftrightarrow f^{-1}f(A)=A for every A\in\text{Codom }f
    Proof:

    \implies:

    x\in A\implies f(x)\in f(A)\implies x\in f^{-1}ff(A) (this is always true). So, A\subseteq f^{-1}f(A)

    Lemma: f(x)\in f(A)\implies x\in A

    Proof: Suppose that x\notin A and let y=f(x). Since y\in f(A) there exists some a\in A such that f(a)=y. But, since x\notin A we see that x\ne a and so f(x)=y=f(a) which condtradicts f's injectivity. The conclusion follows \blacksquare

    Now, then x\in f^{-1}f(A)\implies f(x)\in f(A)\implies x\in A

    so f^{-1}f(A)\subseteq A

    from where it follows that A=ff^{-1}(A)

    \Leftarrow: Let f(x)=f(y) then f(\{x\})=\{f(x)\}=\{f(y)\}=f(\{y\}) and so \{x\}=f^{1}f(\{x\})=f^{-1}f(\{y\})=\{y\} from where it follows that x=y
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    thank

    thank you
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