# Existence of left and right inverses of functions

• prettymidget
In summary: Now let g be not one to one. f,g map X to itself and thus g(x) are elements in X for all x. Since g is not one to one, we can find a x not equal to y such that g(x)=g(y). So we have g(x)=g(y) but f(g(x)) does not equal f(g(y)). So fog is not one to one.
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.

## 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?

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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.

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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?

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.

prettymidget said:
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)).

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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.

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prettymidget said:
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.

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.

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## 1. What is a left inverse of a function?

A left inverse of a function is a function that, when composed with the original function, results in the identity function. In other words, it "undoes" the original function.

## 2. Can a function have more than one left inverse?

No, a function can only have one left inverse. If a function has more than one left inverse, it would result in a contradiction since each left inverse should result in the identity function when composed with the original function.

## 3. What is a right inverse of a function?

A right inverse of a function is a function that, when composed with the original function, results in the identity function. In other words, it "undoes" the original function from the right side.

## 4. Can a function have more than one right inverse?

Yes, a function can have multiple right inverses. However, if a function has a left inverse and a right inverse, then they must be the same function.

## 5. What does it mean if a function has both a left inverse and a right inverse?

If a function has both a left inverse and a right inverse, it is considered to be a bijective function. This means that every element in the domain of the function has a unique output in the range, and vice versa. In other words, the function is both injective (one-to-one) and surjective (onto).

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