# Overview of basic results about images and preimages

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

• This question is also asked in an effort to make search for duplicates easier, since questions like this are asked quite often. For a more detailed discussion see meta. – Martin Sleziak Apr 12 '13 at 18:02
• – Watson Jan 26 '17 at 12:28
• – Watson Jan 26 '17 at 12:30

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. $\newcommand{\Map}{{#1}\colon{#2}\to{#3}}\newcommand{\Img}{{#1}[#2]}\newcommand{\Pre}{{#1}^{-1}[#2]}$

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. $\Img f{A_1\cup A_2}=\Img f{A_1}\cup\Img f{A_2}$.

• The same is true for arbitrary union: $$\Img f{\bigcup_{i\in I} A_i} = \bigcup_{i\in I} \Img f{A_i}.$$
• 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 $\Img f{A_1\cap A_2} = \Img f{A_1} \cap \Img f{A_2}$.

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 $\Img f{\bigcap_{i\in I} A_i} = \bigcap_{i\in I} \Img f{A_i}$.
• Preimages preserve intersection: $\Pre f{B_1\cap B_2}=\Pre f{B_1} \cap \Pre f{B_2}$.
• 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: $\Pre f{B_1\cup B_2}=\Pre f{B_1} \cup \Pre f{B_2}$.

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

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

• $\Img g{\Img fA}=\Img{g\circ f}A$

• $\Pre f{\Pre gC}=\Pre{(g\circ f)}C$

• The amount of duplicates... – Pedro Tamaroff Apr 12 '13 at 18:18
• @user18921 Corrected now. Thanks to wj32. – Martin Sleziak Apr 14 '13 at 17:31
• Yeeees getting this all in one place – Tanner Strunk Aug 28 '17 at 5:06
• Is there a similar list for stronger results when the map is continuous? – plebmatician Dec 20 '18 at 12:12
• @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. – Martin Sleziak Dec 20 '18 at 12:46

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 $T,T'\subseteq Y$.

• $T\supseteq f(f^{-1}(T))$.
• $T\subseteq T' \Rightarrow f^{-1}(T)\subseteq f^{-1}(T')$.
• $f^{-1}(T\cup T')=f^{-1}(T)\cup f^{-1}(T')$.
• $f^{-1}(T\cap T')=f^{-1}(T)\cap f^{-1}(T')$.
• $f^{-1}(T\setminus T')=f^{-1}(T)\setminus f^{-1}(T')$.
• $S\subseteq f^{-1}(f(S))$.
• $S\subseteq S' \Rightarrow f(S)\subseteq f(S')$.
• $f(S\cup S')=f(S)\cup f(S')$.
• $f(S\cap S')\subseteq f(S)\cap f(S')$.
• $f(S\setminus S')\supseteq f(S)\setminus f(S')$.
• $f(S)\cap T=f(S\cap f^{-1}(T))$.
• $f(S)\cup T\supseteq f(S\cup f^{-1}(T))$.
• $S\cap f^{-1}(T)\subseteq f^{-1}(f(S)\cap T)$.
• $S\cup f^{-1}(T)\subseteq f^{-1}(f(S)\cup T)$.
• $(f\circ g)^{-1}(T)=g^{-1}(f^{-1}(T))$.
• $(f\circ g)(R)=f(g(R))$.
• found this while doing a problem from Lee's book. – Tanner Strunk Aug 28 '17 at 5:07

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").

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.