19

Let $f: A \rightarrow B$ be a function. How can we show that for all subsets $S$ of $A$, $S \subseteq f^{-1}(f(S))$? I think this is a pretty simple problem but I'm new to this so I'm confused.

Also, how can we show that $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective?

Martin Sleziak
  • 53,687
GAtechWizard
  • 239

4 Answers4

18

Let $x\in S$. Then $f(x)\in f(S)$, so that $x\in f^{-1}(f(S))$.

Suppose $f$ is injective. Let $x\in f^{-1}(f(S))$. Then $f(x)\in f(S)$, so that $f(x)=f(y)$ for some $y\in S$. It follows that $x=y\in S$.

Suppose $f$ is not injective. Then $f(x)=f(y)=a$ for some $x\neq y$, so that $\{x,y\}\subset f^{-1}(f(\{x\}))$. It follows that $f^{-1}(f(\{x\}))\not\subset \{x\}$.

  • Okay I should have known that. How would I show that $S=f^-1(f(S))$ iff $f$ is one-to-one? – GAtechWizard Dec 18 '12 at 03:22
  • This shows that $S=f^−1(f(S))$ if $f$ is injective but not that $f$ is injective if $S=f^−1(f(S))$. I know you begin by supposing that $f(x) = f(y)$ for some $x,y \in S$ but how do you go about showing that $x = y$? – GAtechWizard Dec 19 '12 at 01:45
  • The reverse direction was proven by contraposition. S=f−1(f(S)) => f injective you can proof this also by f is not injective => S is not f−1(f(S)) – killertoge Feb 26 '21 at 12:49
5

$S \subseteq f^{-1}(f(S)):$ Choose $a\in S.$ To show $a\in f^{-1}(f(S))$ it suffices to show that $\exists$ $a'\in S$ such that $a\in f^{-1}(f(a'))$ i.e. to show $\exists$ $a'\in S$ such that $f(a)=f(a').$ Now take $a=a'.$

$S = f^{-1}(f(S))$ $\forall$ $A \subset S$ $\iff f$ is injective:

  • $\Leftarrow:$ Let $f$ be injective. Choose $s'\in f^{-1}(f(S))\implies f(s')\in f(S)\implies \exists$ $s\in S$ such that $f(s')=f(s)\implies s'=s$ (since $f$ is injective) $\implies s'\in S.$ So $f^{-1}(f(S))\subset S.$ Reverse inclusion has been proved earlier. Therefore $f^{-1}(f(S))= S.$

  • $\Rightarrow:$ Let $f^{-1}(f(S))= S$ $\forall$ $A \subset S.$ Let $f(s_1)=f(s_2)$ for some $s_1,s_2\in S.$ Then $s_1\in f^{-1}(f(\{s_2\})=\{s_2\}\implies s_1=s_2\implies f$ is injective.

Sugata Adhya
  • 3,979
  • The last one it's not correct. We have that $f^{-1}(f({s_2}) = {x : f(x) \in f({s_2})}$. Since $f(s_1) = f(s_2)$, we have that $f^{-1}(f({s_2}) = {s_1, s_2}$, because $f(s_1) \in f({s_2})}$. From this we cannot conclude that $s_1 = s_2$. Subtle, but true. – Paulo Sigiani Apr 28 '23 at 03:00
  • @PauloSigiani I know that this is a 10-year old question... But are you sure that the proof is flawed? By assumption, $f^{-1}\left(f({s_2})\right)={s_1,s_2}={s_2}$. The singleton axiom of set theory guarantees that $s_2=s_1$ as desired. – Dr. Momo Jul 02 '23 at 13:46
  • @Dr.Momo you're right. Think I don't noticed the typo $A\subset S$, when it should be $S \subset A$ and taking $S = {s_2}$. – Paulo Sigiani Jul 02 '23 at 18:20
  • I don't get this step "$f(s′)∈f(S)⟹∃s∈S$ such that $f(s′)=f(s)$". Is it because S is a subset of A? – Omar Feb 05 '24 at 07:44
2

Formal proof (in DC Proof format) at http://dcproof.com/Image-Pre-Image.htm

A straightforward application of definitions except for two "sticking" points:

  1. To prove $\neg x\in S \rightarrow \neg x\in f^{-1}(f(S))$ when $f$ is injective, you must consider two cases: $x\in A$ and $\neg x\in A$.

  2. To prove $\neg\forall S\subseteq A(S=f^{-1}(f(S)))$ when $f$ is not injective, let $x\in A$ and $y\in A$ where $x\ne y$ and $f(x)=f(y)$. Then construct counterexample $S=\{ x\}$.

Dan Christensen
  • 14,529
0

As images and inverse images are compatible with intersection and union, see here or here, this is equivalent with $\{ x \} = f^{-1}(f(x))$ for all $x \in X$ for $f : X \to Y$. Now injectivity is equivalent with $|f^{-1}(x)| \le 1$ for all $x \in X$, but as every element from $f(S)$ with $S \subseteq X$ has at least one element onto which it gets mapped by definition we have $|f^{-1}(f(x))|\ge 1$, taken together we find $|f^{-1}(f(x))| = 1$ or $f^{-1}(f(x)) = \{x\}$. Conversely if this holds, than sureley $|f^{-1}(x)|\le 1$ for every $x \in X$ as either $x \in f(X)$ or not.

Community
  • 1
StefanH
  • 18,086