Exercise 4: Let $f: A \to B, g: B \to C$
be functions. Show that
a) If $f,g$ are both one-to-one, then $g \circ f$ is one-to-one.
b) If $f,g$ are both onto, then $g \circ f$ is onto.
c) If $g \circ f$ is one-to-one, then $f$ is one-to-one.
d) If $g \circ f$ is onto, then $g$ is onto
I have done the second part, Can you guys pls help me with other parts. I have an exam this Wednesday and my professor told that one of these would be coming in exam
b) I got this Assume $f$ and $g$ are onto. We want to prove that $g \circ f$ is onto. Let $c$ be an element of $C$. Since $g$ is onto, there exists $b$ in $B$ such that $g(b) = c$. Now, since $b$ is in $B$ and $f$ is onto, there exists $a$ in $A$ such that $f(a) = b$. So, we get $g(f(a)) = g(b) = c$. This shows that for every $c$ in $C$ there is $a$ in $A$ such that $g(f(a)) = c$, which means $g \circ f$ is onto.
Your second part looks good.
The first one is much like the second one. Use the 1-1-ness of $f$ and $g$ to show that the composition is 1-1. Start with $a_1,a_2\in A$ such that $g(f(a_1))=g(f(a_2))$, and see what you can get from there.
Parts c) and d) are, in my opinion, best proven by contraposition. For c), assume $f$ is not 1-1 (which means there exists $a_1\neq a_2\in A$ such that ...), and show that $g\circ f$ cannot be 1-1 either. For d), assume $g$ is not into (which means there is a $c\in C$ such that ...), and show that $g\circ f$ cannot be onto either.