What you said isn't true -

No, what I said is true, but the example I gave was flawed. Nevertheless it's easy to

fix it: .

This is not a 1-1 function yet ...
the function you give maps two elements to the identity, f(-1)=1=f(1), so what you've written doesn't form a counter-example. Specifically, it doesn't form a counter-example because your map is a homomorphism (as your group is abelian)...

I am not sure what your point is - are you perhaps getting confused with 1-1, which means injection, and a "there exists a 1-1 correspondence between the sets", which means there exists a bijection?

No. My point was that to show in general that a function is 1-1 you must show

that , and it won't help showing that the unit, in case the

set has an algebraic structure that admits one, is the only elt. mapped to the unit.

The famous works in general only for homomorphisms, not

for general functions.

Tonio
Alternatively, counter-example to the comment: `A function is an injection if the only element which maps to the identity is the identity' would be

. This isn't a homomorphism, and so your statement works. In general, the kernel of a homomorphism is trivial if and only if the function is an injection. This is easy to prove, and was what Roninpro was trying to say, and indeed did say when you realise he said ker(f)=e...normal functions don't have kernels...