By Martin Sleziak

2013-04-12 18:02:12 8 Comments

Are there some good overviews of basic facts about images and inverse images of sets under functions?


@Martin Sleziak 2013-04-12 18:02:12

I have no doubt that you there are many useful online resources for these, but many such results are available here at MSE, together with their proofs.

I'll give a list of some basic results about images and preimages links to the posts, which have proofs here at MSE. I am making this CW, so feel free to add more identities and pointers to further useful questions and answers. $\newcommand{\Map}[3]{{#1}\colon{#2}\to{#3}}\newcommand{\Img}[2]{{#1}[#2]}\newcommand{\Pre}[2]{{#1}^{-1}[#2]}$

If $\Map fXY$ is a function and $A\subseteq X$ and $B\subseteq Y$ are some set then the set $$\Img fA=\{f(x); x\in A\}$$ is called the image of the subset $A$ and the set $$\Pre fB=\{x; f(x)\in B\}$$ is called the preimage or inverse image of the subset $B$.

In the other words, we have $$y\in\Img fA \Leftrightarrow (\exists x\in A)f(x)=y$$ and $x\in\Pre fB \Leftrightarrow f(x)\in B$.

In the notation below we always assume $A,A_i\subseteq X$ and $B,B_i\subseteq Y$.

  • $A\subseteq \Pre f{\Img fA}$ and, if $f$ is injective, then $A=\Pre f{\Img fA}$.

see Proving that $C$ is a subset of $f^{-1}[f(C)]$ In fact, the equality is equivalent to the fact that $f$ is injective: Show $S = f^{-1}(f(S))$ for all subsets $S$ iff $f$ is injective or Is $f^{-1}(f(A))=A$ always true? Some counterexamples to the equality can be found in answers to Why $f^{-1}(f(A)) \not= A$

  • $\Img f{\Pre fB}\subseteq B$ and, if $f$ is surjective, then $\Img f{\Pre fB}=B$.

For the first part see: Need help for proving that: $f(f^{-1}(A)) ⊆ A$ For the second part, see Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ or Prove $F(F^{-1}(B)) = B$ for onto function.

  • The operation of taking images and preimages preserves inclusion, i.e. $A_1\subseteq A_2$ implies $\Img f{A_1}\subseteq \Img f{A_2}$ and $B_1\subseteq B_2$ implies $\Pre f{B_1}\subseteq\Pre f{B_2}$

  • Image of union of two sets is the union of their images, i.e. $\Img f{A_1\cup A_2}=\Img f{A_1}\cup\Img f{A_2}$.

See, for example, Prove $f(S \cup T) = f(S) \cup f(T)$ or Functions-Set Theory Proof that $f(C \cup D) = f(C) \cup f(D)$ or How do I prove the following: $f(S\cup T) = f(S) \cup f(T)$ or Maps - question about $f(A \cup B)=f(A) \cup f(B)$ and $ f(A \cap B)=f(A) \cap f(B)$ or Unions and Functions on Sets

  • The same is true for arbitrary union: $$\Img f{\bigcup_{i\in I} A_i} = \bigcup_{i\in I} \Img f{A_i}.$$

See Show that $\bigcup_i f(A_i) = f(\bigcup_i A_i)$

  • In the case of intersection, we only have one inclusion: $$\Img f{A_1\cap A_2}\subseteq \Img f{A_1} \cap \Img f{A_2}.$$ But if $f$ is injective, then $\Img f{A_1\cap A_2} = \Img f{A_1} \cap \Img f{A_2}$.

See, for example Is this a valid proof of $f(S \cap T) \subseteq f(S) \cap f(T)$? or Is this proof correct for : Does $F(A)\cap F(B)\subseteq F(A\cap B) $ for all functions $F$? or Prove $f(S \cap T) \subseteq f(S) \cap f(T)$ or Do we have always $f(A \cap B) = f(A) \cap f(B)$? or Maps - question about $f(A \cup B)=f(A) \cup f(B)$ and $ f(A \cap B)=f(A) \cap f(B)$ or Verifying a proposition on image and preimage: $f(A\cap B)\subseteq f(A)\cap f(B)$ and $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$ or How to prove this? "For all sets A,B⊆D and functions f:D→R, we have f(A∩B)⊆(f(A)∩f(B))." or $f(A \cap B)\subset f(A)\cap f(B)$, and otherwise?

The part about injective functions can be found in Conditions Equivalent to Injectivity or in Proving: $f$ is injective $\Leftrightarrow f(X \cap Y) = f(X) \cap f(Y)$

A counterexample showing that equality is not true in general can be found here: Do we have always $f(A \cap B) = f(A) \cap f(B)$?

In fact, if this equality is true for all subsets $A_1,A_2\subseteq X$, then $f$ must be injective, see To prove mapping f is injective and the other f is bijective

  • Again, the same claims hold for arbitrary intersection: $\Img f{\bigcap_{i\in I} A_i} \subseteq \bigcap_{i\in I} \Img f{A_i}$ and if $f$ is injective, then $\Img f{\bigcap_{i\in I} A_i} = \bigcap_{i\in I} \Img f{A_i}$.

See Proof for Image of Indexed Collection of Sets?

  • Preimages preserve intersection: $\Pre f{B_1\cap B_2}=\Pre f{B_1} \cap \Pre f{B_2}$.

See, for example: What are the strategies I can use to prove $f^{-1}(S \cap T) = f^{-1}(S) \cap f^{-1}(T)$? or how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$ or Verifying a proposition on image and preimage: $f(A\cap B)\subseteq f(A)\cap f(B)$ and $f^{-1}(C\cap D)=f^{-1}(C)\cap f^{-1}(D)$

  • The same is true for arbitrary intersections: $\Pre f{\bigcap_{i\in I}B_i} = \bigcap_{i\in I} \Pre f{B_i}$.

  • Preimages preserve union: $\Pre f{B_1\cup B_2}=\Pre f{B_1} \cup \Pre f{B_2}$.

See: Show that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$

  • Again, this is also true for union of more than just two sets: $$\Pre f{\bigcup_{i\in I}B_i} = \bigcup_{i\in I} \Pre f{B_i}.$$

See: Union of preimages and preimage of union

We can also ask whether image and preimage preserve difference:

  • $\Img fA\setminus\Img fB \subseteq \Img f{A\setminus B}$, but the equality does not hold in general

See Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality. Equality holds for injective functions, see If $f$ is 1-1, prove that $f(A\setminus B) = f(A)\setminus f(B)$. In fact, validity of the equality $\Img fA\setminus\Img fB = \Img f{A\setminus B}$ characterizes injectivity: Does $f(X \setminus A)\subseteq Y\setminus f(A), \forall A\subseteq X$ imply $f$ is injective ?.

  • $\Pre fA\setminus\Pre fB=\Pre f{A\setminus B}$

See Proof of $f^{-1}(B_{1}\setminus B_{2}) = f^{-1}(B_{1})\setminus f^{-1}(B_{2})$. As a consequence of this we get that the preimage of complement is complement of the preimage, see here: How to approach proving $f^{-1}(B\setminus C)=A\setminus f^{-1}(C)$?, Show that for any subset $C\subseteq Y$, one has $f^{-1}(Y\setminus C) = X \setminus f^{-1}(C)$ and Show $f^{-1}(A^c)=(f^{-1}(A))^c$

The image and inverse image for composition of maps can be expressed in a very simple way. For $\Map fXY$, $\Map gYZ$ and $A\subseteq X$, $C\subseteq Z$ we have

  • $\Img g{\Img fA}=\Img{g\circ f}A$

  • $\Pre f{\Pre gC}=\Pre{(g\circ f)}C$

See Prove that if $Z\subseteq Y$, then $(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z)).$

@Pedro Tamaroff 2013-04-12 18:18:49

@Martin Sleziak 2013-04-14 17:31:53

@user18921 Corrected now. Thanks to wj32.

@Tanner Strunk 2017-08-28 05:06:33

Yeeees getting this all in one place

@plebmatician 2018-12-20 12:12:34

Is there a similar list for stronger results when the map is continuous?

@Martin Sleziak 2018-12-20 12:46:44

@plebmatician To be honest, I am not really sure what kind of properties you have in mind. In any case, I have mentioned this in Calculus and analysis chatroom, maybe somebody will notice it there.

@wj32 2013-04-13 23:27:37

This big list is included in Appendix A of Introduction to Topological Manifolds by John M. Lee:

Let $f:X\to Y$ and $g:W\to X$ be maps, and suppose $R\subseteq W$, $S,S'\subseteq X$, and $T,T'\subseteq Y$.

  • $T\supseteq f(f^{-1}(T))$.
  • $T\subseteq T' \Rightarrow f^{-1}(T)\subseteq f^{-1}(T')$.
  • $f^{-1}(T\cup T')=f^{-1}(T)\cup f^{-1}(T')$.
  • $f^{-1}(T\cap T')=f^{-1}(T)\cap f^{-1}(T')$.
  • $f^{-1}(T\setminus T')=f^{-1}(T)\setminus f^{-1}(T')$.
  • $S\subseteq f^{-1}(f(S))$.
  • $S\subseteq S' \Rightarrow f(S)\subseteq f(S')$.
  • $f(S\cup S')=f(S)\cup f(S')$.
  • $f(S\cap S')\subseteq f(S)\cap f(S')$.
  • $f(S\setminus S')\supseteq f(S)\setminus f(S')$.
  • $f(S)\cap T=f(S\cap f^{-1}(T))$.
  • $f(S)\cup T\supseteq f(S\cup f^{-1}(T))$.
  • $S\cap f^{-1}(T)\subseteq f^{-1}(f(S)\cap T)$.
  • $S\cup f^{-1}(T)\subseteq f^{-1}(f(S)\cup T)$.
  • $(f\circ g)^{-1}(T)=g^{-1}(f^{-1}(T))$.
  • $(f\circ g)(R)=f(g(R))$.

@Tanner Strunk 2017-08-28 05:07:07

found this while doing a problem from Lee's book.

@tp1 2013-04-13 15:51:37

Here's interesting application of inverse image:

  1. Given two functions: $ f : R \times R \rightarrow R, g : R \rightarrow 2, g=([n_0..n_1] \mapsto 1)$, the composition of them gives characteristic function $ h : R \times R \rightarrow 2$.
  2. Use equation $h(x,y)=1$ to choose $1$ from $2=\{0,1\}$.
  3. Once chosen, use inverse image to find all $(x,y)$ -pairs: Inverse image $g^{-1}(1)$ gives whole range $[n_0..n_1]$. Inverse image $ f^{-1} ([n_0..n_1]) $ is the interesting one.
  4. Example of how this works is simply by choosing $f = \sqrt{x^2+y^2}$ and $g=([0..5]\mapsto 1)$ to get a (filled) circle with radius 5 from inverse image, distance function f, and simple range function g.
  5. Example2: $f=\sqrt{x^2+y^2}$ and $g=([3..5]\mapsto 1)$ gives you circle with a hole.
  6. Example3: $f=\sqrt{x^2+y^2}$ and $g=([r..r]\mapsto 1)$ gives you solutions to equation $x^2+y^2=r^2$ (with the caveat that the characteristic function is useless)
  7. Key information is that inverse images can give more information about subsets defined via characteristic functions.

@Lord_Farin 2013-04-12 18:10:37

The "online compendium of mathematical proofs" known as ProofWiki has most of these results -- with one or more proofs. (For the record, I am somewhat affiliated with that site.)

Most results regarding images and preimages should be in the Mapping Theory category (look under the "M").

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