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?
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 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? &quot;For all sets A,B⊆D and functions f:D→R, we have f(A∩B)⊆(f(A)∩f(B)).&quot; 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 $(g\circ f)^{-1}(Z)=f^{-1}(g^{-1}(Z)).$
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").