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Theorem:

Suppose $X, Y $ and $C\subseteq X $ are sets and that $f:X\to Y $. Then $C\subseteq f^{-1} (f(C))$.

I don't understand this intuitively. In my mind, $f^{-1}(D) $ is not a mapping, but is a set of all $x\in X $ such that $f (x)\in D $ (Let $D\subseteq Y $). If $f (C) $ is the set that $f^{-1} $ is being evaluated on in the Theorem, I don't see where one could get any extra elements of $X $ from that also map to $f (f^{-1}(C)) $, other than those elements in $C $.

Note: I realized the answer while typing the question up, as often happens.

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There could be some $x_1 \not\in C $ that satisfies $f:x_1\mapsto d\in f (C) $. Then $f^{-1}$ "picks up" $x_1$ after $f (C) $ has been defined.

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Just because you mention intuition, here's a picture highlighting the fact that there's no reason something outside of $C$ can't get sent to a point in the image $f(C)$ as well. preimage of image

As an example, consider the zero map $T\colon \Bbb R^3 \to \Bbb R^3$ that sends each vector $v$ to $T(v) = 0$. Of course each vector in the $xy$-plane $P = \{(x, y, 0): x, y \in \Bbb R\}$ gets sent to $0$ so that $T(P) = \{0\}$. But every other vector does too! Thus $P \subset T^{-1}(T(P)) = T^{-1}(0) = \Bbb R^3$.

I think my drawing was inspired by a graphic I saw that was explaining exact sequences, but I can't remember the book or I'd include that picture as well.

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Write down the bare definitions:

$$f(C):=\left\{\,y\in Y\;:\; \exists\,c\in C\;\;s.t.\;\;f(c)=y\,\right\}$$

and now

$$f^{-1}(f(C)):=\left\{\,x\in X\;:\;\;f(x)\in f(C)\,\right\}$$

But clearly for any $\;c\in C\;$ we have that $\;f(c)\in f(C)\;$ , so $\;C\subset f^{-1}(f(C))\;$ , by definition

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  • $\begingroup$ @ahorn Thank you, you're right, Edited the typo. $\endgroup$ – DonAntonio Apr 8 '16 at 15:32
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Based on definition $f^{-1}(D):=\{x\in X\mid f(x)\in D\}$ we find the (very handsome) rule:$$x\in f^{-1}(D)\iff f(x)\in D$$

If $c\in C$ then $f(c)\in f(C)$ and application of the rule (on $D=f(C)$) gives: $c\in f^{-1}(f(C))$.

Read $f^{-1}(D)$ as "the set of elements that are sent to set $D$".

Then observe that all elements is $C$ are sent to $f(C)$, i.e. $C\subseteq f^{-1}(f(C))$.

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  • $\begingroup$ Thank you. I was more confused about the possibility of a proper inclusion, rather than just an inclusion. $\endgroup$ – ahorn Apr 8 '16 at 15:41
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    $\begingroup$ If e.g. $f:X\to Y$ is a constant function then $f^{-1}(f(C))=X$ for every non-empty $C\subseteq X$. So the inclusion is proper for every non-empty $C$ with $C\neq X$. No such proper inclusions exist if and only if $f$ is injective. $\endgroup$ – drhab Apr 8 '16 at 15:46
  • $\begingroup$ That is a good example. $\endgroup$ – ahorn Apr 8 '16 at 15:52

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