# 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.

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

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

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

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

see Proving that $C$ is a subset of $f^{-1}[f(C)]$
In fact, the equality is equivalent to the fact that $$ff$$ 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 $f^{-1}(f(A)) \not= A$

• $$\Img f{\Pre fB}\subseteq B\Img f{\Pre fB}\subseteq B$$ and, if $$ff$$ is surjective, then $$\Img f{\Pre fB}=B\Img f{\Pre fB}=B$$.
• The operation of taking images and preimages preserves inclusion, i.e. $$A_1\subseteq A_2A_1\subseteq A_2$$ implies $$\Img f{A_1}\subseteq \Img f{A_2}\Img f{A_1}\subseteq \Img f{A_2}$$ and $$B_1\subseteq B_2B_1\subseteq B_2$$ implies $$\Pre f{B_1}\subseteq\Pre f{B_2}\Pre f{B_1}\subseteq\Pre f{B_2}$$

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

• The same is true for arbitrary union: $$\Img f{\bigcup_{i\in I} A_i} = \bigcup_{i\in I} \Img f{A_i}.\Img f{\bigcup_{i\in I} A_i} = \bigcup_{i\in I} \Img f{A_i}.$$
• In the case of intersection, we only have one inclusion: $$\Img f{A_1\cap A_2}\subseteq \Img f{A_1} \cap \Img f{A_2}.\Img f{A_1\cap A_2}\subseteq \Img f{A_1} \cap \Img f{A_2}.$$ But if $$ff$$ is injective, then $$\Img f{A_1\cap A_2} = \Img f{A_1} \cap \Img f{A_2}\Img f{A_1\cap A_2} = \Img f{A_1} \cap \Img f{A_2}$$.

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 $$A_1,A_2\subseteq XA_1,A_2\subseteq X$$, then $$ff$$ must be injective, see To prove mapping f is injective and the other f is bijective

• Again, the same claims hold for arbitrary intersection: $$\Img f{\bigcap_{i\in I} A_i} \subseteq \bigcap_{i\in I} \Img f{A_i}\Img f{\bigcap_{i\in I} A_i} \subseteq \bigcap_{i\in I} \Img f{A_i}$$ and if $$ff$$ is injective, then $$\Img f{\bigcap_{i\in I} A_i} = \bigcap_{i\in I} \Img f{A_i}\Img f{\bigcap_{i\in I} A_i} = \bigcap_{i\in I} \Img f{A_i}$$.
• Preimages preserve intersection: $$\Pre f{B_1\cap B_2}=\Pre f{B_1} \cap \Pre f{B_2}\Pre f{B_1\cap B_2}=\Pre f{B_1} \cap \Pre f{B_2}$$.
• The same is true for arbitrary intersections: $$\Pre f{\bigcap_{i\in I}B_i} = \bigcap_{i\in I} \Pre f{B_i}\Pre f{\bigcap_{i\in I}B_i} = \bigcap_{i\in I} \Pre f{B_i}$$.
• Preimages preserve union: $$\Pre f{B_1\cup B_2}=\Pre f{B_1} \cup \Pre f{B_2}\Pre f{B_1\cup B_2}=\Pre f{B_1} \cup \Pre f{B_2}$$.
• Again, this is also true for union of more than just two sets: $$\Pre f{\bigcup_{i\in I}B_i} = \bigcup_{i\in I} \Pre f{B_i}.\Pre f{\bigcup_{i\in I}B_i} = \bigcup_{i\in I} \Pre f{B_i}.$$

We can also ask whether image and preimage preserve difference:

• $$\Img fA\setminus\Img fB \subseteq \Img f{A\setminus B}\Img fA\setminus\Img fB \subseteq \Img f{A\setminus B}$$, 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 $$\Img fA\setminus\Img fB = \Img f{A\setminus B}\Img fA\setminus\Img fB = \Img f{A\setminus B}$$
characterizes injectivity: Does $f(X \setminus A)\subseteq Y\setminus f(A), \forall A\subseteq X$ imply $f$ is injective ?.

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

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 $$\Map fXY\Map fXY$$, $$\Map gYZ\Map gYZ$$ and $$A\subseteq XA\subseteq X$$, $$C\subseteq ZC\subseteq Z$$ we have

• $$\Img g{\Img fA}=\Img{g\circ f}A\Img g{\Img fA}=\Img{g\circ f}A$$

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