Existence of left and right inverses of functions

[prettymidget]

Homework Statement

Prove or disprove
a) Let f:X---->Y. If f possesses more than 1 left inverse yet has no right inverse, then f has strictly more than 1 left inverse.

b) If f and g are maps from a set X to X and fog is one to one, then f an g are both injective one to one.

Homework Equations

The Attempt at a Solution

a) I came across a counterexample.
f: A -> B where A = {1}, B = {1,2} and f(1)=1b) I know it can be false when f maps X to Y since the only inner function need be one to one , but I am not positive about when it maps to itself. I think the statement becomes true. Amirite?

 

[prettymidget]

Made some typos, here's the right question:

a) If f possesses at least 1 left inverse yet has no right inverse, then f has strictly more than 1 left inverse.

b) If f and g are maps from a set X to X and fog is one to one, then f an g are both one to one.

 

[fzero]

You should show a little more work.

For part a, what condition on f is necessary for the existence of a left inverse? What condition is necessary for the absence of a right inverse?

For part b, if f is not 1-1, can you show whether or not f\circ g is 1-1?

 

[prettymidget]

for b): Let f be not one to one. f,g map X to itself and thus g(x) are elements in X for all x. Since f is not one to one, for some g(x), g(y) with x,y in X, g(x)=g(y) but f(g(x)) does not equal f(g(y)). So fog cannot be one to one.

 

[fzero]

for b): Let f be not one to one. f,g map X to itself and thus g(x) are elements in X for all x. Since f is not one to one, for some g(x), g(y) with x,y in X, g(x)=g(y) but f(g(x)) does not equal f(g(y)). So fog cannot be one to one.

The definition of 1-1 is that if a and b in X, such that a is not equal to b, then f(a) is not equal to f(b). Conversely, if f is not 1-1, then there exist a and b in X, such that a is not equal to b, but f(a)=f(b). So you want to consider that there are g(x) not equal to g(y) but such that f(g(x)) is equal f(g(y)).

 

[prettymidget]

From here since g(x) does not equal g(y), we know x cannot equal y. So we have x,y such that x does not equal y but fog(x)=fog(y). So fog is not one to one.

 

[fzero]

Im confused as to where we go from here.

Well I found a typo in my previous post, I'd written

The definition of 1-1 is that if a and b in X, such that a is not equal to b, then f(a) is not equal to b.

when I meant

The definition of 1-1 is that if a and b in X, such that a is not equal to b, then f(a) is not equal to f(b).

I hope that's not your confusion.

I'd start by assuming that f and fog are 1-1 and that g is not. Then you need to show that there are x and y not equal for which f(g(x))=f(g(y)). But since fog is 1-1, one of our other assumptions must be wrong. Making f not be 1-1 cannot resolve the contradiction, so g must be 1-1. To be complete, we should really repeat the argument with g 1-1 and f not.

 

[prettymidget]

Let f be not one to one. f,g map X to itself and thus g(x) are elements in X for all x.
Since f is not one to one, we can find a g(x) not equal to g(y) such that f(g(x)) is equal to f(g(y)).
Since g(x) does not equal g(y), then x cannot equal y. (This is the step I am a bit unsure of but it seems to work since g(x) and g(y) exist. its equivalent to: If x=y, then g(x)=g(y) which seems very obvious.) So there exists x,y such that x does not equal y but fog(x)=fog(y). So fog cannot be one to one.