Can anyone prove that if a function is invertible, then it must be onto and 1-1?

Question

i'm not sure if this is true, can anyone prove this too me using a legitimate proof

Answer 1

I think your teacher wants you to prove it.

here's a hint to get your started.

To prove one to one:

Your proof should begin.....

"We want to prove....." Now just copy down the definition of one to one from your textbook.

somewhere in that definition is an expression that looks like this:

f(x) = f(y)

[ or f(x1) = f(x2) }

What do you need to prove about x and y ? (it is in the definition you just copied down!)

Now, you know that f is invertible, so what can you do to both sides of this equation?

Answer 2

In fact, a function is invertible if and only if it is one to one and onto.
Let f be a function from A to B and g be its inverse from B to A.
To show that f is one to one, assume not. Then f(x)=f(y) for two distinct x and y. Then g sends f(x) to both x and y, a contradiction.
To show that f is onto, assume not. Then there is some y in B not in the range of f. So g(y) is empty, a contradiction.