## 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 by . That is, define to be the function given by the rule for all . Given , we say that a function is a **left inverse** for if ; and we say that is a **right inverse** for if .

(a) Show that if has a left inverse, is injective; and if has a right inverse, 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 has both a left inverse and a right inverse , then is bijective and ."

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

-----------

SOLUTION

(a)

not injective means that for . The left inverse exists by hypothesis, and so when we apply it to , by definition , and on it is . But, being a function, the left inverse must map to the same single element in , which would mean . We've reached a contradiction in the argument, so is injective.

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

(b)

Many functions are injective but not surjective, say with rule .

(c)

Likewise we can find many functions that are surjective but not injective, say with rule

(d)

More than one left inverse: yes. Here's an example: let with rule . Define by . Another could be by . The important thing to notice is that every element of the domain of is returned to itself after applying and then or .

More than one right inverse: yes. Consider with rule , and define with rule or with rule . Notice that starting from the domain , applying or on zero, then , returns us to zero: the composition of either function does not move the element in its path to and back.

(e)

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

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

Alternatively, by the identity property of equality, . By associativity of composition of functions, . This in turn reduces to , and so .