1.2 Exercise 2

"Let  f : A \rightarrow B   and let  A_i \subset A   and  B_i \subset B   for  i = 0   and  i = 1 .  Show that  f^{-1}   preserves inclusions, unions, intersections, and differences of sets:

(a)  B_0 \subset B_1 \Rightarrow f^{-1}(B_0) \subset f^{-1}(B_1)

(b)  f^{-1}(B_0 \cup B_1) = f^{-1}(B_0) \cup f^{-1}(B_1)

(c)  f^{-1}(B_0 \cap B_1) = f^{-1}(B_0) \cap f^{-1}(B_1)

(d)  f^{-1}(B_0 \setminus B_1) = f^{-1}(B_0) \setminus f^{-1}(B_1)

Show that  f   preserves inclusions and unions only:

(e)  A_0 \subset A_1 \Rightarrow f(A_0) \subset f(A_1)

(f)   f(A_0 \cup A_1) = f(A_0) \cup f(A_1)

(g)  f(A_0 \cap A_1) \subset f(A_0) \cap f(A_1) ; show that equality holds if  f   is injective.

(h)  f(A_0 \setminus A_1) \supset f(A_0) \setminus f(A_1) ; show that equality holds if  f   is injective."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 20.)

NB:  A word about notation.   f^{-1}(B_i) in this context is the preimage of the subset  B_i of  B , and can therefore return a set (i.e., it is not a function), as will be seen presently.

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SOLUTION

(a)

Case  B_0 \subsetneq B_1

Pick  b \in B_0 such that the preimage in the domain  A is defined ( f may not be surjective or even one-to-one).  Apply  f^{-1} on such, and then  f^{-1}(b) \in f^{-1}(B_0) .  Since  b \in B_0 , and it also belongs to  B_1 by (proper) inclusion,  f^{-1}(b) \in f^{-1}(B_1) .  Now pick  b \notin B_0, b \in B_1 , with defined preimage in  A .  Apply  f^{-1} .  Then  f^{-1}(b) \notin B_0, f^{-1}(b) \in B_1 .  Thus  f^{-1}B_0 \subsetneq f^{-1}(B_1) .

Case  B_0 = B_1

It is easy to see that, by arguing similarly,  f^{-1}(B_0) = f^{-1}(B_1) .

Thus  B_0 \subset B_1 \Rightarrow f^{-1}(B_0) \subset f^{-1}(B_1) .  \Box

(b)

Case  f^{-1}(B_0 \cup B_1) \subset f^{-1}(B_0) \cup f^{-1}(B_1)

Pick a  b \in B_0 , for which the preimage is deifned in  A  f^{-1}(b) \in f^{-1}(B_0) \in f^{-1}(B_0) \cup f^{-1}(B_1) .  Pick  b \in B_1  f^{-1}(b) \in f^{-1}(B_1) \in f^{-1}(B_0) \cup f^{-1}(B_1) .  We have covered every  b in  B_0 \cup B_1 , which preimage is  f^{-1}(B_0 \cup B_1) .

Case  f^{-1}(B_0) \cup f^{-1}(B_1) \subset f^{-1}(B_0 \cup B_1)

Pick a  b \in B_0 , well defined in  A .  This element, under  f^{-1} , belongs to  f^{-1}(B_0) and is therefore in  f^{-1}(B_0) \cup f^{-1}(B_1) .  Since  b \in B_0 , it also belongs to  B_0 \cup B_1 , and has a preimage in  f^{-1}(B_0 \cup B_1) .

Pick  b \in B_1 , well defined in  A  f^{-1}(b) \in f^{-1}(B_1) .  Such  b also belongs to  B_0 \cup B_1 , and therefore  f^{-1}(b) \in f^{-1}(B_1 \cup B_0) .

We've now covered every  b in  f^{-1}(B_0) \cup f^{-1}(B_1) , and we are done.

Thus  f^{-1}(B_0 \cup B_1) = f^{-1}(B_0) \cup f^{-1}(B_1) .  \Box

(c)

Case  f^{-1}(B_0 \cap B_1) \subset f^{-1}(B_0) \cap f^{-1}(B_1)

Pick  b \in B_0 \cap B_1 with a defined preimage in  A .  Since it is in  B_0 \cap B_1 , the preimage lies in  f^{-1}(B_0 \cap B_1) .

Such a  b belongs to  B_0 , and so its preimage is in  f^{-1}(B_0) .  It also lies in  B_1 , and so its preimage lies also in  f^{-1}(B_1) .  In other words, its preimage lies in  f^{-1}(B_0) \cap f^{-1}(B_1) .

Case  f^{-1}(B_0) \cap f^{-1}(B_1) \subset f^{-1}(B_0 \cap B_1)

Pick  b \in B_0 , with defined preimage in  f^{-1}(B_0) .  Pick  b' \in B_1 with defined preimage in  f^{-1}(B_1) .  take the intersection  f^{-1}(B_0) \cap f^{-1}(B_1) .

A  b mapping to the intersection has a preimage that maps both to  f^{-1}(B_0) and  f^{-1}(B_1) .  In other words, such a  b \in B_0 and  b \in B_1 , or  B_0 \cap B_1 , so the preimage of  b lies also in  f^{-1}(B_0 \cap B_1) .

Thus  f^{-1}(B_0 \cap B_1) = f^{-1}(B_0) \cap f^{-1}(B_1) .  \Box

(d)

Case  f^{-1}(B_0 \setminus B_1) \subset f^{-1}(B_0) \setminus f^{-1}(B_1)

Pick  b \in B_0, b\notin B_1 , with well defined preimage in  f^{-1}(B_0 \setminus B_1) .  Since  b \in B_0 , its preimage lies in  f^{-1}(B_0) .  But it cannot lie in  B_1 , so therefore its preimage cannot lie in  f^{-1}(B_1) .  In other words, the preimage of such a  b is in  f^{-1}(B_0) \setminus f^{-1}(B_1) .

Case  f^{-1}(B_0) \setminus f^{-1}(B_1) \subset f^{-1}(B_0 \setminus B_1)

Pick a  b with preimage in  f^{-1}(B_0) \setminus f^{-1}(B_1) .  Such a  b has a preimage in  f^{-1}(B_0) , but not in  f^{-1}(B_1) .  This means that such a  b must belong to  B_0 , but not  B_1 :  b \in B_0, b \notin B_1 , or, in other words,  b \in B_0 \setminus B_1 .  Such a  b has a preimage in  f^{-1}(B_0 \setminus B_1) .

Thus  f^{-1}(B_0 \setminus B_1) = f^{-1}(B_0) \setminus f^{-1}(B_1) .  \Box

(e)

Case  A_0 \subsetneq A_1

Pick  a \in A_0 and apply  f .  Such  f(a) \in f(A_0) . By inclusion,  a \in A_1 , and therefore  f(a) \in f(A_1) .  Pick  a \notin A_0 but  a \in A_1 .  Apply  f .  Then  f(a) \notin f(A_0) but  f(a) \in f(A_1) .  Thus  f(A_0) \subsetneq f(A_1) .

Case  A_0 = A_1

Pick  a \in A_0 (and  a \in A_1 ).  Then  f(a) \in f(A_0) and  f(a) \in f(A_1) .  Thus  f(A_0) = f(A_1) .

Therefore  f(A_0) \subset f(A_1) .  \Box

(f)

Same as (b) but arguing form the domain side.

(g)

Case  f(A_0 \cap A_1) \subset f(A_0) \cap f(A_1)

Pick  a \in A_0 \cap A_1 (with image  f(a) \in f(A_0 \cap A_1) ).  Now  a \in A_0 means that  f(a) \in f(A_0) , and  a \in A_1 means  f(a) \in f(A_1) , so such an  a 's image is in  f(A_0) \cap f(A_1) .

Case  f(A_0) \cap f(A_1) \subset f(A_0 \cap A_1) if  f is injective

 f is injective means that there is a unique  a that yields  b \in B , in other words  f(a) = b .  So pick  f(a) \in f(A_0) \cap f(A_1) .  The  a that was picked necessarily belongs to  A_0 in order that  f(a) be in  f(A_0) (i.e., there isn't another  a' outside or inside  A_0 whose image is  f(A_0) , due to injectivity), but it also must belong to  A_1 so that  f(a) be in  f(A_1) (same argument here).  Therefore  a \in A_0 \cap A_1 , and the image of this  a belongs to  f(A_0 \cap A_1) .

If the argument above isn't crystal clear, the contrapositive is also insightful: if  f(A_0) \cap f(A_1) \nsubseteq f(A_0 \cap A_1) ,  f is not injective.  Pick  b \in f(A_0) \cap f(A_1) that is not also in the image  f(A_0 \cap A_1) .  In particular this means  f^{-1}(b) = a' and such  a' \in A_1, a' \notin A_0 , say.  But such a  b \in f(A_0) and  b \in f(A_1) , in other words,  b = f(a) \in f(A_0), f(a)\in f(A_1) and this means that such an  a \in A_0 and  a \in A_1 , or  a \in A_0 \cap A_1 . The only way to resolve this apparent contradiction is by letting  b = f(a') = f(a) , and  f is not injective.  \Box

(h)

Case  f(A_0 \setminus A_1) \supset f(A_0) \setminus f(A_1) , or  f(A_0) \setminus f(A_1) \subset f(A_0 \setminus A_1)

Pick  f(a) \in f(A_0), f(a) \notin f(A_1) .  Since  f(a) belongs to the image  f(A_0) , there exist elements collectively  a \in A_0 , and focus on them (other elements are not of interest because they are not in  A_0 ).  Since  f(a) does not belong to  f(A_1) , these  a \notin A_1 . Such  a exactly belong to  A_0 \setminus A_1 , and therefore  f(a) to  f(A_0 \setminus A_1) .

Next we want to show that if  f is injective,  f(A_0 \setminus A_1) \subset f(A_0) \setminus f(A_1) .  Since all  a s uniquely map into  f(A) , pick any  a \in A_0 \setminus A_1 , and apply  f .  Such will lie in  f(A_0 \setminus A_1) .  Since  a \in A_0 , and  a \notin A_1 , this implies  f(a) \in f(A_0) and  f(a) \notin f(A_1) .  \Box

  1. kris huser
    May 14th, 2012 at 15:08 | #1

    The statement "since f(a) belongs to the image f(A_0) then a belongs to A_0" is false.

    a may not be an element of A_0 but may still be equal to f(a)

  2. May 26th, 2012 at 15:00 | #2

    Sorry, I'm not sure what you mean. You wrote: "a may not be an element of A_0 but may still be equal to f(a)" makes little sense to me... I may be missing something. Perhaps what you meant is that an element outside of A_0 can still map to f(a). If this is the case, then you are right but we are opting for those elements a which definitely belong to A_0, because f(a) \in f(A_0). I'm correcting the wording in the original post a bit to make it clearer.

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