Let g : A -> B and f : B -> C be functions.
Show that if f and g are injective, then f o g is injective.
Show that if f and g are surjective, then f o g is surjective.
i don't know how to proove it
b.
since f is surjective, rng f = C. That is, for any c $\displaystyle \in $C, there exists b $\displaystyle \in $ B such that f(b) = c. Now since g is surjective, there exists an a $\displaystyle \in$ A such that g(a) = b. but then (f o g)(a) = f(g(a)) = f(b) = c, so
f o g is surjective.
For $\displaystyle f\circ g$ to be injective $\displaystyle f(g(x))=f(g(y))\implies x=y$. Assume $\displaystyle f(g(x))=f(g(y))$. Since $\displaystyle f$ is injective...
Spoiler:
For $\displaystyle f\circ g$ to be surjective, $\displaystyle \forall~y\in C$, $\displaystyle \exists~x\in A$ such that $\displaystyle f(g(x))=y$. Let $\displaystyle y\in C$. Because $\displaystyle f$ is surjective...
Spoiler: