# More on the homology functor

Following from this question, I am having trouble constructing the Homology functor on chain maps. Let me first sum up what I managed to do.

Let $$\textbf{A}$$ be an abelian category, and let $$C_\bullet,D_\bullet\in\textbf{Ch}_\bullet(\textbf{A})$$ be two chain complexes, and let $$f:C_\bullet\to D_\bullet$$ be a chain map.

# 1. Construction of $$H_n$$ on objects

For each $$n\in\mathbb{Z}$$, there is a monomorphism $$\text{Im}(\partial_{n+1})\to\ker(\partial_n)$$. Indeed, from the mono/epi decomposition of $$\partial_{n+1}$$, one gets a map $$g:C_{n+1}\to\text{Im}(\partial_{n+1})$$ :

Now $$g$$ is epic, because $$\widehat{\partial_{n+1}}$$ is an iso and because $$\text{Coim}(\partial_{n+1})$$ is epic, as a cokernel. At last, $$0=\partial_n\partial_{n+1}=\partial_ncg$$, and since $$g$$ is epic, we get $$\partial_nc=0$$. Therefore, from the universal property for $$\ker(\partial_n)$$, we get :

Now $$\bar{c}$$ is monic, for if $$x:\ker(\partial_n)\to\bullet$$ is such that $$\bar{c}x=0$$, then $$k\bar{c}x=cx=0$$, i.e. $$x=0$$ since $$c$$ is monic (it is an image, which in turn, is a kernel). This allows us to define : $$H_n(C)=\text{coker}(\bar{c}):\ker(\partial_n)\to H_n(C).$$

# 2. Definition of $$H_n$$ on chain maps

The composite $$f_n\ker(\partial_n^{(C)})$$ is such that (using the definition of a chain map to "commute" $$f_n$$ and $$\partial_n$$) : $$\partial_n^{(D)}f_n\ker(\partial_n^{(C)})=f_{n-1}\partial_n^{(C)}\ker(\partial_n^{(C)})=0.$$

From the universal property for $$\ker(\partial_n^{(D)})$$, this gives rise to a unique morphism :

At last, we are left with a canonically-given diagram :

# 3. How to conclude ?

What I need to show is that there is a unique morphism $$H_n(C)\to H_n(D)$$ making the previous diagram commutative. How to do so ? From the linked question above, the answer suggests using universal properties from (co)kernels : how to ?

I tried using $$\text{coker}(\bar{c})$$ :

In order to have the desired map, I need to prove that $$H_n(D)\hat{f_n}\bar{c}_n^{(C)}=0$$ : how to do so ?

# 4. Conclusion

Let's follow Jacob FG's indication.

Let's use the universal property of kernels for $$\text{coker}(\partial_{n+1}^{(D)})$$. We have the following commutative diagram :

From this, we get : $$\text{coker}(\partial_{n+1}^{(D)})f_n\partial_{n+1}^{(C)}=\text{coker}(\partial_{n+1}^{(D)})\partial_{n+1}^{(D)}f_{n+1}=0.$$

Now recall the mono/epi canonical decomposition in any abelian category : $$\partial_{n+1}^{(D)}=\text{Im}(\partial_{n+1}^{(D)})\overline{\partial_{n+1}^{(D)}}\text{Coim}(\partial_{n+1}^{(D)}),$$

where $$\overline{\partial_{n+1}^{(D)}}\text{Coim}(\partial_{n+1}^{(D)})$$ is epic, as the composition of an epi with an iso. From this, we obtain finally : $$\text{coker}(\partial_{n+1}^{(D)})f_n\text{Im}(\partial_{n+1}^{(C)})=0,$$

which in turn, yields :

Now, we wish to prove that the following diagram is commutative :

Finally, recall the three relations, obtained from the previous diagrams :

$$\begin{cases}\text{Im}(\partial_{n+1}^{(D)})\tilde{f_n}=f_n\text{Im}(\partial_{n+1}^{(C)})\\\ker(\partial_n^{(D)})\hat{f_n}=f_n\ker(\partial_n^{(C)})\\\ker(\partial_n^{(D)})\bar{c}_n^{(D)}=\text{Im}(\partial_{n+1}^{(D)})\end{cases}.$$

From there relations, we get : $$\ker(\partial_n^{(D)})\bar{c}_n^{(D)}\tilde{f_n}=\text{Im}(\partial_{n+1}^{(D)})\tilde{f_n}=f_n\text{Im}(\partial_{n+1}^{(C)})=f_n\ker(\partial_n^{(C)})\bar{c}_n^{(C)}=\ker(\partial_n^{(D)})\hat{f_n}\bar{c}_n^{(C)}.$$

Finally, from monomorphicity of $$\ker(\partial_n^{(D)})$$, we obtain the desired relation, thus commutativity of the diagram.

This finally yields the desired relation : $$H_n(D)\hat{f_n}\bar{c}_n^{(C)}=H_n(D)\bar{c}_n^{(D)}\tilde{f_n}=0,$$

since $$H_n(D)=\ker(\bar{c}_n^{(D)})$$ by definition. This provides us with a unique arrow $$H_n(C)\to H_n(D)$$ in the cokernel diagram, and this arrow is what we wanted.

You should use that $$f$$ is a morphism of chain complexes to prove that the following diagram commutes:
Then you get that $$H_n(D)\hat{f}_n\bar{c}_n^{(C)} = H_n(D)\bar{c}_n^{(D)}\hat{f}_{n} = 0$$.