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Let $f: A \rightarrow B$ be given. Prove:

(a) For each subset $X \subset A, X \subset f^{-1}(f(X)).$

At first I didn't understand why $X$ wouldn't just be equal to $f^{-1}(f(X))$, but I guess that is the difference between an inverse image and an inverse function, right? For instance if I had $A = \{1,2,3\}, B = \{1,2\}, X = \{1,2\}, W = \{2,3\}$, and $f(x) = 1$ then $X, W \subset A$, but $f^{-1}(f(X)) = \{1,2,3\} \neq X$ and of course $X \subset f^{-1}(f(X))$, right?

So am I understanding at least that much correctly? My problem is that I don't know what I am actually supposed to write down to prove this in general, could someone please help me with that?

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This is almost exclusively about definitions. What is the definition of $f(X)$? What about $f^{-1}(Y)$ for $Y \subset B$? Work your way from the "inside" out. – cardinal Mar 13 '11 at 16:39
up vote 1 down vote accepted

Let $Y\subset B$. To say that $x\in f^{-1}(Y)$ is the same as saying that $f(x)\in Y$. Now, if $x\in X$, then $f(x)\in f(X)$, and hence $x\in f^{-1}(f(X))$.This proves that $X\subset f^{-1}(f(X))$. The inclusion may be strict if $f$ is not injective. For instance, if $A=B=\mathbb{R}$ the real numbers, $f(x)=x^2$ and $X=[0,\infty)$, then $f^{-1}(f(X))=\mathbb{R}$. This happens because $f$ is not injective: for each $y>0$ there are two values of $x$ such thet $f(x)=x^2=y$: namely, $\pm\sqrt{y}$. Of these two values only the positive one is in $X$.

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About the strictness of inclusion: equality holds iff $f(x)=f(a)\Rightarrow x=a$ for all $x\in X,a\in A$. In particular this is satisfied when $f$ is injective. – wildildildlife Mar 13 '11 at 17:01

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