Injectivity of Function Composition: A Closer Look at Left and Right Inverses

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Homework Statement

a) Prove or disprove: If f:X---->Yhas at least one left inverse g:Y---->X but has no right inverse, then f has more than one such left inverse.

b) Prove or disprove: If f and g are maps from a set X to X and fog is injective, then f an g are both injective. (fog being function composition).

Homework Equations

The Attempt at a Solution

a) I think it is true.

Assume f only has one such inverse, i.e. g is unique.
If f has no right inverse, there exists no map h such that f(h(a))=a for all a in X.
g(f(a))=a for all a in X.

b) False, found a counterexample. the inner function need not be injective. Still stuck on a though.

 

[prettymidget]

No one? :(

 

[prettymidget]

I hate bumping threads but I'm getting desperate. I found that my counterexample for b does not work because I forgot that f and g need to map X into itself, and now I actually think b may be true.