More homework help. Given the function $f:A \to B$. Let $C$ be a subset of $A$ and let $D$ be a subset of $B$.

Prove that:

$C$ is a subset of $f^{-1}[f(C)]$

So I have to show that every element of $C$ is in the set $f^{-1}[f(C)]$

I know that $f(C)$ is the image of $C$ in $B$ and that $f^{-1}[f(C)]$ is the pre-image of $f(C)$ into $A$. Where I'm stuck is how to use all of this information to show/prove that $C$ is indeed a subset.

Do I start with an arbitrary element (hey, let's call it $x$) of $C$? and then show that $f^{-1}[f(x)]$ is $x$? I could use a little direction here... Thanks.


Since you want to show that $C\subseteq f^{-1}\big[f[C]\big]$, yes, you should start with an arbitrary $x\in C$ and try to show that $x\in f^{-1}\big[f[C]\big]$. You cannot reasonably hope to show that $f^{-1}\big[f[\{x\}]\big]=x$, however: there’s no reason to think that $f$ is $1$-$1$, so there may be many points in $A$ that $f$ sends to the place it sends $x$.

Let $x\in C$ be arbitrary. For convenience let $E=f[C]\subseteq B$. Now what elements of $A$ belong to the set $f^{-1}\big[f[C]\big]=f^{-1}[E]$? By definition $f^{-1}[E]=\{a\in A:f(a)\in E\}$. Is it true that $f(x)\in E$? If so, $x\in f^{-1}[E]=f^{-1}\big[f[C]\big]$, and you’ll have shown that $C\subseteq f^{-1}\big[f[C]\big]$.

  • $\begingroup$ So it was all about using the definition of the preimage along the way to prove my assumption. It is always so clear in hindsight. Hopefully this gets easier before I get to Abstract Algebra and Real Analysis... $\endgroup$ – Ben Anderson Apr 1 '13 at 5:11
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    $\begingroup$ @Ben: You’ll find that a very large fraction of the more elementary exercises really do just boil down to working through the relevant definitions. You generally have to get a bit deeper into a subject before you encounter problems that require much more than that. $\endgroup$ – Brian M. Scott Apr 1 '13 at 5:13
  • $\begingroup$ @BrianM.Scott: I have added my answer to a question recently marked as a duplicate of this one. If you feel it is too similar to yours, I will remove it. $\endgroup$ – robjohn Sep 1 '15 at 21:49
  • $\begingroup$ @robjohn: Both the difference in style and the inclusion of the example make it worth keeping. (Besides, this made me catch a bad typo in my answer!) $\endgroup$ – Brian M. Scott Sep 2 '15 at 4:26

Copied from my answer to I am issues with proving the following problem: $f^{-1}(f(A)) ⊃ A$, which was closed as a duplicate of this question just before I posted it.

$$ f^{-1}\left(f(A)\right)=\left\{x:f(x)\in f(A)\right\}\tag{1} $$ Note that if $x\in A$, then $f(x)\in f(A)$, and by $(1)$, $x\in f^{-1}\left(f(A)\right)$. Therefore, by definition, we have $$ A\subset f^{-1}\left(f(A)\right)\tag{2} $$ However, if $f$ is not injective, then $f^{-1}\left(f(A)\right)$ may indeed contain elements not present in $A$; for example let $f:\mathbb{Z}\mapsto\mathbb{Z}$ be defined by $$ f(x)=\left\lfloor\frac x2\right\rfloor\tag{3} $$ and let $A$ be the set of even integers. Then $f(A)=\mathbb{Z}$ and $$ A\subsetneq\mathbb{Z}=f^{-1}\left(f(A)\right)\tag{4} $$


The problem can be reduced to one-line proof in the following way.

Let $f : Y \leftarrow X$ denote a function.

The definition of inverse images says that for all $y : Y$ and all $x : X,$ we have:

$$f^*(y) \ni x \iff y = f(x)$$

This can be rewritten:

$$f^*(y) \supseteq \{x\} \iff \{y\} \supseteq \{f(x)\}$$

This can be used to prove:

Proposition. For all subsets $B$ of $Y$ and all subsets $A$ of $X$, we have:

$$f^*(B) \supseteq A \iff B \supseteq f_*(A)$$

Once you've proved this, your problem becomes a one-line proof:

$$f^{*}(f_*(A)) \supseteq A \iff f_*(A) \supseteq f_*(A)$$


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