For simplicity let's say I am working in the category of chain complexes of $R$ modules. I have a chain map $i:C\rightarrow D$, which I know to be a weak equivalence (i.e. $i_*$ is an isomorphism of homologies), and a monomorphism. Since $i$ is a monomorphism, I obtain the short exact sequence:
$$0\xrightarrow{\ker i} C\xrightarrow{i} D\xrightarrow{\text{coker }i} K\longrightarrow 0$$
which gives rise to the long exact sequence of homologies:
$$\cdots \xrightarrow{d}H_n(C)\xrightarrow{(i_*)_n}H_n(D)\xrightarrow{(\text{coker } i_*)_n}H_n(K)\xrightarrow{d}H_{n-1}(C)\xrightarrow{(i_*)_{n-1}}H_{n-1}(D)\cdots$$
I want to argue that $H_n(K)$ of is identically zero for all $n$. For context, I am trying to prove that the category of chain complexes is a model category. Specifically, I am trying to show that fibrations have the right lifting property with respect to trivial cofibrations, and came across the need for $K$ to be acyclic.
Anyways, here is the argument:
I know that $(i_*)_n$ is an isomorphism, and since the sequence is exact, I know that $\text{im } (i_*)_n=H_n(D)=\ker (\text{coker }i_*)_n$, so $(\text{coker }i_*)_n$ is the zero map. This implies that the image of $(\text{coker }i_*)_n$ is $0$, so the kernel of the right most $d$ is zero, however, the kernel of $(i_*)_{n-1}$ is also zero, so the image of the right most $d$ is zero as well. I then conclude that since $d$ has zero kernel, but also sends everything to $0$, that $H_n(K)$ must be zero, and this holds for all $n$.
Does this argument make sense? I am specifically worried about about the last step, regarding $d$ be injective, but also the zero map implying that $H_n(K)$ must be zero, but I feel like that has to be true...
Any help would be greatly appreciated.