Let $f:A\rightarrow B$, so $g:{\mathcal P}(B)\rightarrow \mathcal{P}(A)$.
By $g(B)=$$''$$f^{-1}$$''$$(B)$, prove that $f$ is injective iff $g$ is surjective.
I'd appreciate any help
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There are two $;f$'s there...which one is what? – DonAntonio Dec 14 '13 at 13:58
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1What is $P(B)$? Power set of $B$? – Dec 14 '13 at 13:59
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yes, it's the power set – gazok Dec 14 '13 at 14:03
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then you should specify that and you could remove $"f^{-1}"$ and just write $f^{-1}$ and you may want to rewrite $g(B)=f^{-1}(B)$ as $g(M):=f^{-1}(M)$ for any $M\subset B$ – Dec 14 '13 at 14:05
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Assume that $f$ is injective and let $X$ be a subset of $A$. Then $f(X)$ is a subset of $B$ and I claim that $g(f(X)) = X$. Indeed, $g(f(X)) = f^{-1}(f(X))$. For any map $f$, we always have $X \subset f^{-1}(f(X))$. Moreover, the inverse inclusion is true if $f$ is injective. Indeed, if $y$ is in $f^{-1}(f(X))$, then $f(y) \in f(X)$ and by injectivity, $y \in X$.
Conversely, assume that $g$ is surjective. Consider $x_1, x_2$ in $A$ such that $f(x_1) = f(x_2)$. There is a $Y \subset B$ such that $g(Y) = \{x_1\}$, this means that $f(x_1)$ is in $Y$. But then, $f(x_2) = f(x_1)$ is also in $Y$ which shows that $x_2 \in g(Y)$. Thus, $x_1 = x_2$ and this proves the injectivity of $f$.
Jeremy Daniel
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If $f$ is a function from $A$ to $B$ and $X$ is a subset of $A$, the direct image of $X$ by $f$ is $f(X) := {y \in B: \exists x \in X : f(x) = y}$. You should learn it again. – Jeremy Daniel Dec 14 '13 at 15:20
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$f$ is injective iff $\forall E\in P(A)$, $f^{-1}(f(E)) = E$. So if $f$ is injective, then $g$ is surjective with $g(f(E)) = E$ for all $E \in P(A)$.
If $g$ is surjective and if $f(x) = f(y) = a$ with $x,y\in A$. Then we have $g(X) = \{x\}$ and $g(Y) = \{y\}$ for some $X, Y\in P(B)$; $f(x) \in X$ and $f(y) \in Y$; hence $a \in X$ and $a\in Y$; hence $\{x\} = \{y\} = g(\{a\})$. This implies $f$ is injective.
Du Phan
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I assume your function $g\colon P(B)\to P(A)$ is defined by $$ g(Y)=\{x\in A: f(x)\in X\} $$ for $Y\subseteq B$ (that is, $Y\in P(B)$).
Assume $g$ is surjective and suppose $f(x_1)=f(x_2)$; set $y=f(x_1)=f(x_2)$. Then there esists $Y\in P(B)$ such that $g(Y)=\{x_1\}$. In particular, $y=f(x_1)\in Y$. But then $y=f(x_2)\in Y$ as well. Can you conclude that $x_1=x_2$?
Assume now that $f$ is injective. For $X\in P(A)$, consider $Y=\{f(x):x\in A\}$. Can you say that $g(Y)=X$?
egreg
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