Exercise 1.1.2, Weibel

Question

From Weibel's An introduction to homological algebra:

Exercise 1.1.2

Show that a morphism $u:C\rightarrow D$ of chain complexes sends boundaries to boundaries and cycles to cycles, hence maps $H_n(C)\rightarrow H_n(D)$. Prove that each $H_n$ is a functor from $\mathbf{Ch}(\mathbf{mod}-R)$ to $\mathbf{mod}-R$.

I'm having trouble with the second part of this exercise, and I'm pretty new to category theory. What am I supposed to show? Is it enough to show that:

1. $H_n(C)\in\mathbf{mod}-R$ for every chain $C\in\mathbf{Ch}(\mathbf{mod}-R)$.
2. For each morphism $u:C\rightarrow D$ in $\mathbf{Ch}(\mathbf{mod}-R)$, $H_n(u):H_n(C)\rightarrow H_n(D)$ is a morphism in $\mathbf{mod}-R$.
3. $H_n(\text{id}_C)=\text{id}_{H_n(C)}$, for every $C\in\mathbf{Ch}(\mathbf{mod}-R)$.
4. $H_n(f\circ g)=H_n(f)\circ H_n(g)$, for every morphism $f,g$ in $\mathbf{mod}-R$.

Attempt

1. OK!

2. Consider the commutative diagram below $\require{AMScd}$\begin{CD}\ldots @>>>C_{n+1}@>{d_{n+1}}>>C_n @>{d_n}>> C_{n-1}@>>>\ldots\\ ~ @V{u}VV @V{u}VV @V{u}VV\\ \ldots @>>>D_{n+1}@>{d'_{n+1}}>>D_n @>{d'_n}>> D_{n-1}@>>>\ldots\end{CD} I will show:

   • Boundaries gets mapped to boundaries: Let $x\in\text{im}(d_{n+1})$, then $u\circ d_{n+1}(x)=d'_{n+1}\circ u(x)$, so $u(x)\in\text{im}(d'_{n+1})$.

   • Cycles gets mapped to cycles: Let $x\in\ker(d_n)$, then $d'_n\circ u(x)=u\circ d_n(x)=0$, so $u(x)\in\ker(d'_n)$.

   • The map $H_n(u):H_n(C)\rightarrow H_n(D)$ is well-defined: Let $x\equiv x'\in H_n(C)$, then I need to show $u(x)\equiv u(x')\in H_n(D)$. This amounts to show that $u(x)-u(x')\in\text{im}(d'_{n+1})$. Since $x\equiv x'$, we have that $x-x'\in\text{im}(d_{n+1})$. Therefore let $d_{n+1}(y)=x-x'$, since the diagram commutes we have $$u(x)-u(x')=u\circ d_{n+1}(y)=d'_{n+1}\circ u(y)$$

   • Linearity of $H_n(u)$: This is immediate, since $u$ is linear.

3. $\require{AMScd}$ \begin{CD} \ldots @>>>H_{n+1}(B)@>{H_{n+1}(f)}>>H_n(B) @>{H_n(f)}>> H_{n-1}(B)@>>>\ldots\\ ~ @V{H_{n+1}(u)}VV @V{H_n(u)}VV @V{H_{n-1}(u)}VV\\ @>>>H_{n+1}(C)@>{H_{n+1}(d)}>>H_n(C) @>{H_n(d)}>> H_{n-1}(C)@>>>\ldots\\ ~ @V{H_{n+1}(\text{id}_C)}VV @V{H_n(\text{id}_C)}VV @V{H_{n-1}(\text{id}_C)}VV\\ \ldots @>>>H_{n+1}(C)@>{H_{n+1}(g)}>>H_n(C) @>{H_n(g)}>> H_{n-1}(C)@>>>\ldots\\ ~ @V{H_{n+1}(v)}VV @V{H_n(v)}VV @V{H_{n-1}(v)}VV\\ @>>>H_{n+1}(D)@>{H_{n+1}(h)}>>H_n(D) @>{H_n(h)}>> H_{n-1}(D)@>>>\ldots \end{CD} To show that $H_n(\text{id}_C)=\text{id}_{H_n(C)}$, do I have to show that $H_n(\text{id}_C)\circ H_{n}(u)=H_{n}(u)$ and $H_n(v)\circ H_n(\text{id}_C)=H_n(v)$? If that is the case then

   • Let $x\in H_n(B)$, then $H_n(\text{id}_C)\circ H_n(u)(x)=H_n(\text{id}_C)(u(x))=H_n(u)(x)$

   • Let $x\in H_n(C)$, then $H_n(v)\circ H_n(\text{id}_C)(x)=H_n(v)(x)=H_n(v)(x)$

4. Let $\require{AMSCD}$\begin{CD}H_n(C)@>{H_n(g)}>>H_n(D)@>{H_n(f)}>>H_n(E)\end{CD} be morphisms between chain complexes.

   • Let $x\in H_n(C)$, then $H_n(f\circ g)$ sends $x\mapsto (f\circ g)(x)\in H_n(E)$.
   • Let $x\in H_n(C)$, then $H_n(g)$ sends $x\mapsto g(x)\in H_n(D)$, and $H_n(f)$ sends $g(x)\mapsto (f\circ g)(x)\in H_n(E)$

Is this correct? I feel that I draw arrows without knowing what is going on.

Answer 1

Answer to you very first question: yes, that's all you have to show by definition of a functor.

After your edit everything seems correct. On part 3, however, it is enough to show that $H_n(id_C)(x)=x$ for every $x\in H_n(C)$, because we're in the category of $R$-modules and that's how one defines the identity there. What you did is the general procedure for an arbitrary category so it's fine.

Finally, you missed something on part 2. You did show that it is a well-defined map, but that's not enough to be a morphism of $R$-modules. This will easily follow from linearity of the boundary maps.

If you're still not sure what's going one make sure you do everything step by step using the definitions on each step and you'll see everything matches the arrows.