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1.2 Exercise 6

January 26th, 2009 No comments

This was a really easy problem, that ends the section on functions in Munkres's text.  The next fifteen problems deal with relations, and I am finding them immensely interesting.

"Let  f : \mathbb R \rightarrow \mathbb R be the function  f(x) = x^3 - x .  By restricting the domain and range of  f appropriately, obtain from  f a bijective function  g .  Draw the graphs of  g and  g^{-1} . (There are several possible choices for  g .) "

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 21.)

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SOLUTION  

As an example,  g: (-\infty, -1) \rightarrow \mathbb{R}^{-} with rule  g(x) = x^3 -x is bijective.

1.2 Exercise 5

January 23rd, 2009 No comments

After working on these problems all week, I'm not sure I can keep up 1 problem a day.  I'll try to post as many as possible in the span of a week, however.  I'm definitely taking a break over the weekend!

I really like this problem because it was an eye-opener when I first encountered it on the subtleties of inverse mappings.  I've seen it several places, it is a classic exercise on functions.  Here's the problem as taken from Munkres's text:

"In general, let us denote the identity function for a set  C by  i_C .  That is, define  i_C : C \rightarrow C to be the function given by the rule  i_C(x) = x for all  x \in C .  Given  f : A \rightarrow B , we say that a function  g : B \rightarrow A is a left inverse for  f if  g \circ f = i_A ; and we say that  h : B \rightarrow A is a right inverse for  f if  f \circ h = i_B .  

(a) Show that if  f has a left inverse,  f is injective; and if  f has a right inverse,  f is surjective. 

(b) Give an example of a function that has a left inverse but no right inverse. 

(c) Give an example of a function that has a right inverse but no left inverse. 

(d) Can a function have more than one left inverse? More than one right inverse? 

(e) Show that if  f has both a left inverse  g and a right inverse  h , then  f is bijective and  g=h=f^{-1} ."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 21.)

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SOLUTION  

(a)  

 f not injective means that  f(a) = f(a') for  a \neq a' .  The left inverse exists by hypothesis, and so when we apply it to  f(a) , by definition  (g \circ f)(a) = a , and on  f(a') it is  (g \circ f)(a') = a' .  But, being a function, the left inverse must map  f(a) = f(a') = b \in B to the same single element in  A , which would mean  a = a' . We've reached a contradiction in the argument, so  f is injective.

Suppose  f is not surjective.  This means there exists a subset of elements  B_{*} \subset B that do not map back to themselves via  f , but then this contradicts the fact that all elements of  B come back to themselves after applying  h and then  f . Thus,  f must therefore be surjective. 

(b)  

Many functions are injective but not surjective, say  f: \mathbb{R}^{+} \rightarrow \mathbb{R} with rule  \{(x,x) \vert x \in \mathbb{R}\} .  

(c)  

Likewise we can find many functions that are surjective but not injective, say  f : [0, \pi] \rightarrow [0,1] with rule  \{ (x, sin(x)) \vert x \in \mathbb{R}\}  

(d)

More than one left inverse: yes.  Here's an example: let  f : \{0,1\} \rightarrow \{0, 1, 2\} with rule  \{ (0,1) ; (1,2) \} .  Define  g : \{0,1, 2\} \rightarrow \{0, 1 \} by  \{ (0,0) ; (1,0) ; (2,1) \} .  Another could be  g': \{0,1, 2\} \rightarrow \{0, 1 \} by  \{(0,1) ; (1,0) ; (2,1)\} .  The important thing to notice is that every element of the domain of  f is returned to itself after applying  f and then  g or  g' .

More than one right inverse: yes.  Consider  f : \{0,1\} \rightarrow \{0\} with rule  \{ (0,0) ; (1,0) \} , and define  h: \{0\} \rightarrow \{0,1\} with rule  \{ (0,0) \} or  h': \{0\} \rightarrow \{0,1\} with rule  \{ (0,1) \} .  Notice that starting from the domain  B , applying  h or  h' on zero, then  f , returns us to zero: the composition of either function does not move the element in its path to  A and back.  

(e)  

Since  f has both a left inverse and a right inverse, it is both injective and surjective, and hence bijective.

To show that  g = h , pick any element  b \in B and apply  h  h is defined on all  B (in fact on the image of  f by surjectivity, and specifically never outside it).  Furthermore, since  f is injective, there is only one way to reach the element  h(b) = a \in A .  Next pick the same element  b \in B and apply  g .  Since  f is surjective, there is no risk of picking a  b that is defined to map back to  A in the same way as another  b' -not-in-the-image-of- f under  g . There is only one way  g can map that  b into  A by the injectivity of  f , so  g(b) = a . Since  h and  g both map to the same element in  A for all  b \in B ,  g = h .  Call this the inverse of  f , and denote it by  f^{-1} .  (Note this is different than the preimage of f which uses the same symbol).

Alternatively, by the identity property of equality,  g \circ f \circ h = g \circ f \circ h .  By associativity of composition of functions,  (g \circ f) \circ h = g \circ (f \circ h) .  This in turn reduces to  (i_A) \circ h = g \circ (i_B) , and so  h = g .

1.2 Exercise 4

January 22nd, 2009 2 comments

I've skipped Exercise 3 because it was basically a repeat of Exercise 2, only requiring induction.  There are plenty of induction-related proofs in the next section, but perhaps I'll return to it if the more interesting problems are depleted.

"Let  f : A \rightarrow B and  g : B \rightarrow C .

(a) If  C_0 \subset C , show that  (g \circ f)^{-1}(C_0) = f^{-1}g^{-1}(C_0) .

(b) If  f and  g are injective, show that  g \circ f is injective.

(c) If  g \circ f is injective, what can you say about injectivity of  f and  g ?

(d) If  f and  g are surjective, show that  g \circ f is surjective.

(e)  If  g \circ f is surjective, what can you say about surjectivity of  f and  g ?

(f) Summarize your answers to (b)-(e) in the form of a theorem."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 21.)

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SOLUTION

(a)

Pick a  c \in C_0 with defined preimage in  A under  (g \circ f)^{-1} .  In other words, a  c such that  (g \circ f)^{-1}(c) \in (g \circ f)^{-1}(C_0) (we need a defined preimage since  (g \circ f) may not be surjective).  Now, such a  c must have arrived at set  B , and therefore must have a defined preimage in  B , as  g^{-1}(c) .  In turn this  g^{-1}(c) must arrive in  A via  f^{-1} , as  (f^{-1} \circ g^{-1}) (c) \in (f^{-1} \circ g^{-1})(C_0) .  This shows the first inclusion, that  (g \circ f)^{-1}(C_0) \subset (f^{-1} \circ g^{-1})(C_0) .

Pick a  c with defined preimage in  B and in  A via  g^{-1} and  f^{-1} respectively. I.e., pick  c \in C_0 so that  (f^{-1} \circ g^{-1})(c) \in (f^{-1} \circ g^{-1})(C_0) .  This  c will have a preimage in  A via  (g \circ f)^{-1} , since if it were undefined in this inverse map, it would have been undefined at either the map  g^{-1} or the map  f^{-1} , but this is not the case.  In other words,  (g \circ f)^{-1}(c) \in (g \circ f)^{-1}(C_0) .  This shows the second inclusion, that  (g \circ f)^{-1}(C_0) \supset (f^{-1} \circ g^{-1})(C_0) .

Finally,  (g \circ f)^{-1}(C_0) = (f^{-1} \circ g^{-1})(C_0)  \Box .

(b)

Suppose not (indirect proof), that  g \circ f is not injective.  This means that there exists  a and  a' \in A such that  (g \circ f)(a) = (g \circ f)(a') .  But this means that we arrived at such element of  C by landing on the same element at  B , meaning  f(a) = f(a') (which contradicts the injectivity of  f ), or by a different element of  B , but no question at the same element after applying the map  g , so that  g(f(a)) = g(f(a')) .  But this contradicts the injectivity of  g .  We must have been wrong that  (g \circ f)(a) is not injective, and so it must be   \Box .

(c)

Rephrasing the question, we want to find out   g \circ f injective  \Rightarrow  f injective and, or, neither  g injective.  Suppose  f is not injective (indirect proof), and so there are elements  a and  a' \in A that map to the same element in  B , as  f(a) = f(a') .  Next apply  g , so that  g(f(a)) = g(f(a')) .  We started from two different elements in  A , and we've now shown the map  g \circ f cannot be injective, a contradiction.   f must therefore be injective.  Next suppose  g is not injective, and pick  b \neq b' \in B , so that  g(b) = g(b') .  If such  b, b' are in the image of  f , and since  f is injective by the previous argument, then  b = f(a) and  b' = f(a') with  a \neq a' \in A .  Starting from two different elements of  A , the map  g \circ f cannot be injective, a contradiction. However, if, say,  b is not in the image of  f ,  g \circ f can still be one-to-one.  Thus we cannot tell that  g is or is not injective.

(d)

Arguing this by contradiction (indirect proof) makes the proof a very simple task.  If the map  g \circ f is not surjective, either  g   is not surjective because all the elements of  B do not hit all elements of  C , or if  g is surjective, then  f is not, hitting elements of  B for which the image under  g is a proper subset of  C . (Note that those elements not in the image of  f in  B  could still hit all the rest of  C , making  g   surjective)  \Box .

(e)

Rephrasing, we want to find out  g \circ f surjective  \Rightarrow  f   surjective and, or, neither  g   surjective.  It's easiest if we start with the map  g .  If  g is not surjective, there is no way  g \circ f can be, because we've got to stop at  B and take the map  g on our way to  C .  Thus,  g \circ f   surjective  \Rightarrow    g surjective.  f   not surjective means only the image of  f in  B will map to  C under composition of functions. But then either such image under  g hits all of  C (no contradiction), or it doesn't (contradiction), we cannot tell.  So the behavior of  f is not particularly illuminating.  \Box .

(f)

Summarizing:

 f  and  g  injective  \Rightarrow    g \circ f  injective

 f  and  g  surjective  \Rightarrow    g \circ f  surjective

 g \circ f injective  \Rightarrow  f injective

 g \circ f  surjective  \Rightarrow    g  surjective

1.2 Exercise 2

January 21st, 2009 2 comments

"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

1.2 Exercise 1

January 20th, 2009 No comments

"Let   f : A \rightarrow B . Let   A_0 \subset A   and   B_0 \subset B .

(a) Show that   A_0 \subset f^{-1}(f(A_0))   and that equality holds if   f   is injective.

(b) Show that   f(f^{-1}(B_0)) \subset B_0   and that equality holds if   f   is surjective."

(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   f   is not injective  

 f   not injective over the domain  A   implies there exist at least two   a \in A  for some   b \in f(A) . More to the point, for some  b \in f(A)  f^{-1}(b) = \{a_1, a_2, ...\} where the cardinality of  \{a_1, a_2, ...\}   is at least 2.

 \mathit{I.}   Pick   A_0 \subset A   in such a way that   A_0 \cap \{a_1, a_2, ...\} contains at least one element but not all.  Apply   f on  A_0 , and then  b \in f(A_0) .  Apply  f^{-1} on  f(A_0) , but in so doing we are sure to apply   f^{-1} on  b .  Since  f^{-1}(b) = \{a_1, a_2, ...\} for which   A_0 contains some, but not all elements (the others which lie in   A ),  f^{-1}(f(A_0)) \supset A_0 .  Here it is important to remember that we did not restrict   f   or   r (the rule of f) to  A_0 , just  A to  A_0 .

*  \mathit{II.}   Pick   A_0 \subset A   such that   A_0 \cap \{a_1, a_2, ...\} = \{\} .  Then (a multivalued)   b \notin f(A_0)   and   f   is "locally" one-to-one.  In this case,   f^{-1}(f(A_0)) = A_0 .  (See   f   is injective below).

*  \mathit{III.}   Pick   A_0 \subset A   such that   A_0 \cap \{a_1, a_2, ...\} = \{a_1, a_2, ...\} .  In this case,    f^{-1}(f(A_0)) = A_0   also.

*  \mathit{IV.}   Lastly, pick   A_0 = A .     f^{-1}(f(A_0)) = A_0   in this case too because the function returns the whole domain. 

NOTES: Starred statements (*) make sense because the inclusion "  \subset " is not a proper inclusion (we need to be careful with notation and follow Munkres's convention), and therefore both sets on either side of the symbol may be equal (in other words,   f^{-1}(f(A_0)) = A_0 ). For   \mathit{III, IV} , any multivalued   b \in B_0   falls via   f^{-1}   on   A_0 .    

Case  f is injective: If   f   is injective, then   f^{-1}(f(A_0))=A_0     

Since there doesn't exist a   b \in f(A) \vert f^{-1}(b) = \{a_1, a_2, ...\}   for   \Vert \{a_1, a_2, ...\} \Vert \geq 2 , this implies that   \forall b \in f(A)  f^{-1}(b) = a   and this   a   is unique.  

Well, surjectivity is established vacuously because   f(A_0)   is defined as that set in the range   B   which is made to correspond with all elements of   A_0   (call it   B_0 ).

Thus   f   becomes bijective:   \exists  only one   b \in B_0   that corresponds to only one   a\in A_0 , and vice versa by definition of the rule of   f .  So there exists an inverse function to   f , and   f(a) = b \iff a = f^{-1}(b) .  Thus

 f(A_0) = B_0

 f^{-1}(f(A_0)) = f^{-1}(B_0)  

 f^{-1}(f(A_0)) = A_0      \Box .

(b)    

Case  f   is not surjective

  f   not surjective over   B   means there is at least one   b \in B   such that   f^{-1}   is not defined (is null).  Let's denote this set by   \{b_1, b_2, ... \}   with cardinality   \geq 1 .

 \mathit{I.}   Pick   B_0 \subset B   in such a way that   B_0 \cap \{b_1, b_2, ...\}   contains at least one element and   B_0 \setminus \{b_1, b_2, ...\}   likewise contains at least one element.  Upon applying   f^{-1}   on   B_0 , we obtain elements in   A_0   (sometimes for some elements   b \in B_0   there may be many   a \in A_0   that correspond, since   f^{-1}   is not one-to-one), except for those elements   B_0 \cap \{b_1, b_2, ...\}   which by definition are undefined in   A .  Applying   f   on such   A_0   returns those elements    B_0 \setminus \{b_1, b_2, ...\} .  It seems clear that    B_0 \setminus \{b_1, b_2, ...\} \subset B_0   by definition of difference of sets, since    B_0 \setminus \{b_1, b_2, ...\} = \{b \vert b \in B_0 and  b \notin \{b_1, b_2, ...\}\} .  Therefore   f(f^{-1}(B_0)) \subset B_0 .  

  \mathit{II.}   As above, except   B_0 \setminus \{b_1, b_2, ...\}   can be empty.  Then the argument is as before, with   B_0 \setminus \{b_1, b_2, ...\} = \{\} = f(f^{-1}(B_0))   and   \{\} \subset B_0 is vacuously true.    

Case  f  is surjective

This argument is likewise as before, except the set of undefined values in   B   is null.  Then   f(f^{-1}(B_0)) = B_0 \setminus \{\} = B_0 , and the equality is established.    \Box