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 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
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? “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}.

See: Proving set equalities

  • 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 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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