1.2 Exercise 5

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.)




 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. 


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


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}\}  


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.  


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 .

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