Monomorphisms and epimorphisms in the category of chain complexes Let $\mathsf{C}$ be an abelian category and $\mathsf{Comp(C)}$ its category of chain complexes. Suppose that $f\colon (C,d)\to (C',d')$ is a monomorphism in $\mathsf{Comp(C)}$. I want to prove that each $f_n\colon C_n\to C'_n$ is a monomorphism in $\mathsf{C}$. Similarly, if $f$ is an epimorphism, I want to prove that each $f_n$ is an epimorphism (I'm not sure it can be proven by duality from assuming the monomorphism case, in particular, I don't think $\mathsf{Comp(C^{op})} = \mathsf{Comp(C)^{op}}$).
If $g,h\colon X\rightrightarrows C_n$ such that $f_n\circ g = f_n\circ h$, to show that $g = h$, we need to construct a complex $(C'',d'')$ and a morphisms $g',h'\colon (C'',d'') \rightrightarrows (C,d)$ such that $g'_n = g, h'_n = h$ and $g'\circ f = h'\circ f$. But I can't guees what $(C'',d'')$ and $g',h'$ should be.
 A: The category of chain complexes is isomorphic to a full subcategory of the category of functors $\mathbb{Z}\to \mathsf{C}$ (where $\Bbb Z$ is zeen as a thin category), closed under limits and colimits. In fact it is even isomorphic to a category of enriched functors (see this question) into $\mathsf{C}$.
So all limits and colimits of chain complexes are constructed "pointwise", which implies that mono/epi-morphisms are pointwise mono/epimorphisms, and also that the category is abelian.
A: This answer assumes you know that your category of complexes is abelian.
Given a chain map $f_*\colon (C_*, d)\to (C_*', d')$, we may consider the complex $(K_*, \delta)$, where $K_n\ker(f_n)$ and $\delta_n$ is the restriction of $d_n$ to $K_n$. To see this, it suffices to show that if $f_n(x)=0$, then $d_n(x)\in\ker(f_{n-1})$, but this follows from the fact that $f_*$ is a chain map: $d'_n\circ f_n = f_{n-1}\circ d_n$. Thus, we have the morphisms $\delta_n\colon K_n\to K_{n-1}$, and they satisfy $\delta_{n-1}\circ\delta_n$. You can then show that $(K_*, \delta_*)$ is a kernel of $f_*$. Then, like in any abelian category, $f_*$ is a monomorphism if and only if its kernel is trivial. Note that triviality does not mean contractibility, but that $K_n=0$ for each $n$. Thus, $f_*$ is a monomorphism if and only if $K_n=0$ for each $n$, but this is equivalent to $f_n$ being a monomorphism.
For epimorphisms, the idea is similar. Here, however, we construct a complex using cokernels. Again, the fact that $f_*$ is a chain map induces a differential between the consequent cokernels: $d'_n(\text{im}(f_n))\subseteq \text{im}(f_{n-1}\circ d_n)\subseteq\text{im}(f_{n-1})$. Now $f_*$ is an epimorphism if and only if the cokernel complex (which is a cokernel of $f_*$!) is trivial, but then the cokernels of all $f_n$ are trivial, and so each $f_n$ is an epimorphism.
