f: A -> B is 1-1 iff whenever f(x) = f(y), then x = y.
Let f: A -> B and let D subset of A. The restriction of f to D is the function
f|D = {(x, y): y = f(x) and x element D}.
The theorem makes sense. For example, f: R -> R where f(x) = x is 1-1. If we restirct the domain to D subset of R F: D -> R where f(x) = x is still 1-1.
I just don't know how to formally prove this.
Thanks
Let A and B be non-empty sets.
And f:A --> B be a one-to-one map.
Let D be a non-trivial subset of A.
Then g: D --- > B defined as g(x)=f(x).
We need to show that g(a)=g(b) implies a=b.
Let us assume that it fails, i.e. their this collapsing. Meaning we can find "a" and "b" distinct such that, f(a)=f(b). But then that means f(a)=f(b). But that cannot be because f is one-to-one. Thus, a contradiction.