Bijective functions and finding a composition function fg

[leej72]

Homework Statement

Give a genuinely new (i.e. not discussed in class or in the book or in tutorial) example of: two sets X and Y , and two functions f : X →Y and g : Y → X, such that the composition g ◦ f is the identity function 1X : X → X, but neither f nor g are bijective. (Reminders: if f : X→ Y and g : Y → Z are two functions, then the composition g ◦ f is a function from X to Z defined by (g◦f)(x):=g(f(x))forallx∈X.The identity function1X :X→X is defined by1X(x):=x for all x ∈ X.)

Homework Equations

The Attempt at a Solution

I was just wondering since f and g are inverses of each other, wouldn't they have to be bijective in order for their composition to be the identity function X --> X? In general, I'm just very confused that if you put in a certain input, you would get the same input as an output. Wouldn't the set size of f also be equal to the set size of g?

 

[Chaos2009]

Well, suppose |X| is strictly less than |Y| and both finite. Then f wouldn't be bijective since it cannot be onto. Similarly g couldn't be bijective since it cannot be one-to-one. However, if we restrict our attention to only the subset of |Y| that is hit by f (say Z := f(X)), then f and g could be bijective on that set. This is the only thing I can think of right now.