Let f: A-B and g: B-C. If (g o f) is 1-1, then f is 1-1. Is this proof right?

Suppose f(x1) = f(x2). Since (g o f)(x1) = (g o f)(x2) = x1 = x2, f is one to one.

Question

Let f: A->B and g: B->C. If (g o f) is 1-1, then f is 1-1. Is this proof right?

Suppose f(x1) = f(x2). Since (g o f)(x1) = (g o f)(x2) => x1 = x2, f is one to one.

anonymous · Answer

The proof doesn't look right.

anonymous · Answer

how come?

anonymous · Answer

What if $f(x_1)\neq f(x_2)$?

anonymous · Answer

then f wouldn't be a function right?

anonymous · Answer

What?  Why not?

anonymous · Answer

it's because (g o f) is one to one. So by definition, if (g o f)(x1) = (g o f)(x2), then x1 = f2

anonymous · Answer

typo then x1 = x2

anonymous · Answer

The proof is just so short I wasn't sure if it's right.

anonymous · Answer

Wait, whoops.  It doesn't matter if $f(x_1)\neq f(x_2)$.  What matters is that you assumed $(g\circ f)(x_1) = (g\circ f)(x_2)$, but what is the justification for that?

anonymous · Answer

isn't it just part of the definition?

anonymous · Answer

Oh, I think I see it now.  Yeah the proof looks right.

anonymous · Answer

you just gave me a heart attack :D

anonymous · Answer

I keep reading it backwards.

anonymous · Answer

lol yeah i had trouble memorizing the order of composite functions quite a long time too