Kernel of chain map in $\textbf{Ch}(\mathcal{A})$ with $\mathcal{A}$ an abelian category (Weibel 1.2.3)

Question

I’m reasonably new to Homological algebra and category theory. I’m working through Weibel and I’m getting stuck on exercise 1.2.3, and theorem 1.2.3. If $\mathcal{A}$ is an abelian category I want to show that $\textbf{Ch}(\mathcal{A})$ is an abelian category.

$\textbf{My attempt}$.
If $f$ is a chain map in $\textbf{Ch}(\mathcal{A})$, I want to show that $f$ has a kernel and cokernel.

I’ve shown that in $\textbf{Ch}(R-\textbf{mod})$ the complexes $\{ker(f)\}$ is the kernel of $f$ similarly for its cokernel. But I am not sure how this works in the general case of chain complexes over an abelian category, when $ker(f)$ becomes a map.

I’ve tried…

$$ \begin{array}{cccccc} A_{n+1} & & A_{n} & & A_{n-1}\\ \quad \downarrow_{ker(f_{n+1})} & & \quad \downarrow_{ker(f_n)} & & \quad \downarrow_{ker(f_{n-1})} \\ C_{n+1} & \xrightarrow{d_{n+1}} & C_n & \xrightarrow{d_n} & C_{n-1}\\ \downarrow_{f_{n+1}} & & \downarrow_{f_{n}} & & \downarrow_{f_{n-1}} \\ D_{n+1} & \xrightarrow{d_{n+1}} & D_n & \xrightarrow{d_n} & D_{n-1} \end{array} $$

And tried to see if these maps are what is required, but how can I define the sequence of $\{A_n\}$ to be a chain complex, do I put the zero map between each of them? And if so, how can I be sure then that $ker(f)$ is a chain map??

Given that I cant even show that kernels and cokernels exist I have very little hope of showing that every monic is the kernel of its cokernel etc...

Answer 1

Since $$A_n\to C_n\to C_{n-1}\to D_{n-1}=A_n\to C_n\to D_n\to D_{n-1}=0,$$ a unique morphism $d_n\colon A_n\to A_{n-1}$ is induced from $A_n\to C_n\to C_{n-1}$. In general, the zero would not make the diagram commute.

Next, $$\begin{align}A_{n+1}\to A_n\to A_{n-1}\to C_{n-1} &= A_{n+1}\to A_n\to C_n\to C_{n-1} \\ &=A_{n+1}\to C_{n+1}\to C_n\to C_{n-1} \\ &= 0, \end{align}$$ hence already $A_{n+1}\to A_n\to A_{n-1}=0$. This makes $A_\bullet$ a chain complex.

The same type of (here, element-less) diagram chasing will help you show that $A_\bullet$ (together with $\ker f_\bullet$) has the universal property of a kernel; that the dual argument works for cokernels; and so on.