If $G,H$ are abelian groups, $f_1:G\to H$ and $f_2:H \to G$ are group homomorphisms such that $f_1f_2=\text{id}_H$, then $G \cong \ker(f_1) \oplus H$
This question is HW from http://www.math.buffalo.edu/~badzioch/MTH619/Lecture_Notes_files/MTH619_week6.pdf (not my HW, but not sure if I can post this)
My attempt:
As $f_1 \circ f_2 = \text{id}_H$, $f_2$ is injective and $f_1$ is surjective. So all elements $H$ can be expressed as $f_1(g)$ for some $g \in G$ and $f_1(g_1)=f_1(g_2)$ implies $g_1=g_2$.
I tried the map $\pi:\ker(f_1) \oplus H \to G$ by $\pi (k,h) = kf_2(h)$. Then $f_1 \pi(k,h)=f_1(k)h=h$, and I was able to show injectivity. But I don’t know how to show surjectivity.
Injective: \begin{align*} \pi(k_1,h_1)=\pi(k_2,h_2) & \implies k_1f_2(h_1)=k_2f_2(h_2) \\ & \implies h_1=h_2 \text{ by applying $f_1$}\\ & \implies k_1=k_2 \text{ by cancellation on $f_2(h)$} \end{align*}
Surjective: want $g=kh$ for all $g \in G$ for some $k \in \ker(f_1), h \in H$