Im taking abstract structure of math and am having a hard time with function proofs (set and subset proofs made so much more sense). For class tomorrow we are asked to attempt to prove theorem 3.2 and I have no idea how to complete it. (really I'm just looking for more ideas, hints, or guidance)
Theorem 3.2: Let f: A->B and g: B->C be functions. (1) If gf is one-to-one then f is one-to-one (2) If gf is onto, then g is onto. (3) If f and g are both bijections then g*f is a bijection.
What I have so far:
Proof: Let f: A->B and g: B->C be functions. (1) Assume g*f is one-to-one, then there exists an x and y in A, such that f(x)=f(y), implying x=y.
I'M SO LOST HOW TO CONTINUE FROM THERE. HELP PLEASE.
Best Answer
1) You know that $g \circ f(x) = g \circ f(y)$ implies $x=y$ since $g \circ f$ is injective. Now suppose $f(x)=f(y)$. What does that tell you about $g \circ f(x)$ and $g \circ f(y)$ (think definition of function)?
2) Let $y \in C$. Then, there is some $x \in A$ so that $g \circ f(x)=y$. If we want $x_{0} \in B$ so that $g(x_{0})=y$ what would be a good candidate?
3) If $f:A \to B$ and $g: B \to C$ are both bijective, then they are injective and surjective. So, $f(x)=f(y)$ implies $x=y$ for $x,y \in A$ and $g(x_{0})=g(y_{0})$ implies $x_{0}=y_{0}$ for $x_{0},y_{0} \in B$. What can we then say if $g \circ f(x) = g \circ f(y)$ (hint: $f(x),f(y) \in B$)? Finally, let $z \in C$. We know we can find $y \in B$ such that $g(y)=z$. But $f$ is also surjective so what can you use that to conclude?