I am trying to solve Weibel Exercise 1.4.4 which asks us to show that a split chain complex is homotopy equivalent to its homology.
Setup:
A. Given a chain complex $C$, its homology is the chain complex $H_*(C):=\{H_n(C),0\}$ ie the differentials are all zero.
B. A chain complex being split means there is a collection of maps $\{s_n:C_{n}\to C_{n+1}\}$ st $d_n=d_ns_{n-1}d_n$ for all $n$.
Lemma 1 (an exercise in Weibel): $C$ is split $\implies$ $C_n\cong B_n\oplus H_n(C)\oplus B_{n-1}$, where $B_n:=\text{Im}(d_{n+1})$ for the differential $d_{n+1}:C_{n+1}\to C_n$.
C. Two chain maps $f,g:C\to D$ are homotopic if there is a collection of maps $\{s_n:C_{n}\to D_{n+1}\}$ st $f_n-g_n=s_{n-1}d_n+d_{n+1}s_n$.
D. Two chain complexes $C$ and $D$ are homotopy equivalent if there is a chain map $f:C\to D$ for which there is a chain map $g:D\to C$ st $fg$ is homotopic to $id_D$ and $gf$ is homotopic to $id_C$.
Problem: Let $C$ be a split chain complex. We want to show that there is some homotopy equivalence $C\to H_*(C)$.
Here is my candidate(BTW I am not at all sure whether these candidates are the right choice or not): Using Lemma 1 (stated above), since $C$ is split, I have an isomorphism $\varphi_n:C_n\to B_n\oplus H_n(C)\oplus B_{n-1}$ for each $n$. To keep it simple, I will drop the subscripts from morphism names. Consider, $$\pi_2\varphi: C_n\to B_n\oplus H_n(C)\oplus B_{n-1}\to H_n(C),$$ and its potential(?) homotopy inverse, $$\varphi^{-1}\iota_2:H_n(C)\to B_n\oplus H_n(C)\oplus B_{n-1}\to C_n.$$ Here $\pi_{2}$ is the canonical projection map (here I am using that in abelian categories finite products and finite co-products co-incide) $B_n\oplus H_n(C)\oplus B_{n-1}\to H_n(C)$ and $\iota_2$ is the canonical inclusion map $H_n(C)\to B_n\oplus H_n(C)\oplus B_{n-1}$.
Lemma 2 (see here, Remark 12.3.6): Given a binary product (or binary co-product) $x_1\oplus x_2$ then $\pi_1\iota_1=id_{x}$, $\pi_2\iota_2=id_{x_2}$, $\pi_1\iota_2=0$, $\pi_2\iota_1=0$ and $\iota_1\pi_1+\iota_2\pi_2=id_{x_1\oplus x_2}$. One can extend this to a product of three terms to get similar equations, $\pi_i\iota_i=id_{x_i}$, $\pi_i\iota_{j\neq i}=0$, and $\sum_{i=1}^{3} \iota_i\pi_i=id_{x_1\oplus x_2\oplus x_3}$.
$\pi_2\varphi\varphi^{-1}\iota_2=id_{H_n(C)}$ from Lemma 2.
Lastly, $\varphi^{-1}\iota_2\pi_2\varphi=\varphi^{-1}(id-\iota_1\pi_1-\iota_3\pi_3)\varphi=id_{C_n}-\varphi^{-1}(\iota_1\pi_1+\iota_3\pi_3)\varphi$.
That is, $id_{C_n}-\varphi^{-1}\iota_2\pi_2\varphi=\varphi^{-1}(\iota_1\pi_1+\iota_3\pi_3)\varphi$.
If I could somehow show that $\varphi^{-1}(\iota_1\pi_1+\iota_3\pi_3)\varphi$ is of the form $ds+sd$, then I would be done.
Any suggestions and/or corrections will be greatly helpful :))