Greatest Hits on Quiz 3
Problem 1
The case distinction in y is not the same as x. It should be y\ge 0 and y<0.
All the set-up in a proof may be similar each time, but it is not optional. You start your proof with "Let x\in \mathbb R. Let y=f(x) (the hypothesis)." And you continue from there.
Note that this amounts to only half of the proof of g being f's inverse, despite what some of you claimed. A full proof would also include the converse of the statement proved.
- A popular mistake was to prove the converse of what was asked--see also the previous point.
Problem 2
- You're not supposed to apply any rules in writing the dual or the set theory variant of an equality.
Problem 3
Proof of 3b)
To show: If g\circ f=\iota_A, then f is one-to-one.
Proof: Let g\circ f=\iota_A. To show: f is one-to-one, which is equivalent to \forall x,y\in A, f(x)=f(y) \rightarrow x=y. (Not to be confused with \forall x,y\in A, x=y \rightarrow f(x)=f(y), which is true for any function.)
So let x,y\in A and let f(x)=f(y). Then obviously g(f(x))=g(f(y)), but by g\circ f=\iota_A, that means \iota_A(x)=\iota_A(y), which in turn means x=y, which is what we needed to show.