Are there some good overviews of basic facts about images and inverse images of sets under functions?
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5This 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](http://meta.math.stackexchange.com/questions/3967/wouldorganizerpostsbeusefulwelcomehere). – Martin Sleziak Apr 12 '13 at 18:02

Related: http://math.stackexchange.com/questions/746123/provethatff1fxfx, http://math.stackexchange.com/questions/1061602 – Watson Jan 26 '17 at 12:28

Related: http://math.stackexchange.com/questions/608707, http://math.stackexchange.com/questions/834689 – Watson Jan 26 '17 at 12:30
4 Answers
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 FunctionsSet 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
or The preimage of a subset
 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 $A,B\subseteq D$ and functions $f:D\mapsto R$, we have $f(A\cap B)\subseteq(f(A)\cap f(B))$."
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: Inverse image of a union equals the union of the inverse images and 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 11, 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 Show that $(g \circ f)^{1}(C) = g^{1}(f^{1}(C)).$ and Prove that if $Z\subseteq Y$, then $(g\circ f)^{1}(Z)=f^{1}(g^{1}(Z)).$
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2[The amount of duplicates...](http://i.huffpost.com/gen/210699/thumbs/sTHERENTISTOODAMNHIGHJIMMYMCMILLANlarge.jpg) – Pedro Apr 12 '13 at 18:18

@user18921 Corrected now. Thanks to [wj32](http://math.stackexchange.com/posts/359694/revisions). – Martin Sleziak Apr 14 '13 at 17:31


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](https://chat.stackexchange.com/transcript/14150/2018/12/20), maybe somebody will notice it there. – Martin Sleziak Dec 20 '18 at 12:46

2I love the fact almost all of this can be deduced immediately from the adjunctions $f[]\vdash f^{1}[] \vdash f_{*}[]$. – Jackozee Hakkiuz May 05 '19 at 21:25
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 $T,T'\subseteq Y$.
 $T\supseteq f(f^{1}(T))$.
 $T\subseteq T' \Rightarrow f^{1}(T)\subseteq f^{1}(T')$.
 $f^{1}(T\cup T')=f^{1}(T)\cup f^{1}(T')$.
 $f^{1}(T\cap T')=f^{1}(T)\cap f^{1}(T')$.
 $f^{1}(T\setminus T')=f^{1}(T)\setminus f^{1}(T')$.
 $S\subseteq f^{1}(f(S))$.
 $S\subseteq S' \Rightarrow f(S)\subseteq f(S')$.
 $f(S\cup S')=f(S)\cup f(S')$.
 $f(S\cap S')\subseteq f(S)\cap f(S')$.
 $f(S\setminus S')\supseteq f(S)\setminus f(S')$.
 $f(S)\cap T=f(S\cap f^{1}(T))$.
 $f(S)\cup T\supseteq f(S\cup f^{1}(T))$.
 $S\cap f^{1}(T)\subseteq f^{1}(f(S)\cap T)$.
 $S\cup f^{1}(T)\subseteq f^{1}(f(S)\cup T)$.
 $(f\circ g)^{1}(T)=g^{1}(f^{1}(T))$.
 $(f\circ g)(R)=f(g(R))$.
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Very nice! There is a similar theoremsexercises on the first chapter of Tom Apostol's mathematical analysis. – C.F.G Sep 15 '20 at 18:57
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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Here's interesting application of inverse image:
 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$.
 Use equation $h(x,y)=1$ to choose $1$ from $2=\{0,1\}$.
 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.
 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.
 Example2: $f=\sqrt{x^2+y^2}$ and $g=([3..5]\mapsto 1)$ gives you circle with a hole.
 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)
 Key information is that inverse images can give more information about subsets defined via characteristic functions.
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