the inverse image commutes with set operations


Theorem. Let f be a mapping from X to Y. If {Bi}iI is a (possibly uncountable) collectionMathworldPlanetmath of subsets in Y, then the following relationsMathworldPlanetmathPlanetmath hold for the inverse imagePlanetmathPlanetmath:

  1. (1)

    f-1(iIBi)=iIf-1(Bi)

  2. (2)

    f-1(iIBi)=iIf-1(Bi)

If A and B are subsets in Y, then we also have:

  1. (3)

    For the set complementPlanetmathPlanetmath,

    (f-1(A))=f-1(A).
  2. (4)

    For the set differenceMathworldPlanetmath,

    f-1(AB)=f-1(A)f-1(B).
  3. (5)

    For the symmetric differenceMathworldPlanetmathPlanetmath,

    f-1(AB)=f-1(A)f-1(B).

Proof. For part (1), we have

f-1(iIBi) = {xXf(x)iIBi}
= {xXf(x)Bifor someiI}
= iI{xXf(x)Bi}
= iIf-1(Bi).

Similarly, for part (2), we have

f-1(iIBi) = {xXf(x)iIBi}
= {xXf(x)Bifor alliI}
= iI{xXf(x)Bi}
= iIf-1(Bi).

For the set complement, suppose xf-1(A). This is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to f(x)A, or f(x)A, which is equivalent to xf-1(A). Since the set difference AB can be written as ABc, part (4) follows from parts (2) and (3). Similarly, since AB=(AB)(BA), part (5) follows from parts (1) and (4).

Title the inverse image commutes with set operations
Canonical name TheInverseImageCommutesWithSetOperations
Date of creation 2013-03-22 13:35:24
Last modified on 2013-03-22 13:35:24
Owner matte (1858)
Last modified by matte (1858)
Numerical id 11
Author matte (1858)
Entry type Proof
Classification msc 03E20
Related topic SetDifference