I am trying to get a better understanding of homology from a Category Theory perspective. Apparently, the following is a well-known fact:
Fact. Let $R$ be a ring. Then, the inclusion functor $i: _{R}Mod \rightarrow _{R}Chain_{\geq}0$ is right adjoint to $H_{0}: _{R}Chain_{\geq}0 \rightarrow _{R}Mod$. (There is an analogous statement with cochains and the inclusion functor being a left adjoint.)
(Here, $_{R}Mod$ means "the category of $R$-modules", and $_{R}Chain_{\geq}0$ means "the category of $R$-chain complexes with $M_{i} = 0$ for negative indexes".)
I have two questions.
(First question.) How is (explicitly) the isomorphism between specific modules and chains?
I want to prove the fact. To do so, I have to take an arbitrary chain $X = \{... \rightarrow X_{1} \rightarrow X_{0} \rightarrow X_{-1} \rightarrow ...\}$ (with maps $d_{-1}, d_{0}, d_{1}$ and so on) and an arbitrary $R$-module $Y$, and I have to show that there is an isomorphism between these sets: $ Hom (i(Y), X) = \{$ arrows from $Y$ to $X_{0} \}$ and $Hom (Y, H_{0}(X)) = \{$ module homomorphisms $Y \rightarrow \frac{Ker d_{0}}{Im d_{1}} \}$. What is the isomorphism given by?
(Second question.) Why is the statement not true when we replace $_{R}Chain_{\geq}0$ with $_{R}Chain$?
I am guessing that this is not true because I have not been able to find a more general version of Fact. I was wondering if someone could give me some insights on what goes wrong when we try to remove the $\geq0$ restriction, or a counterexample that shows it in a clear manner.
You want to show that $i$ is the right adjoint to $H_0$, i.e. you have to prove that $\mathrm{Chain}(R)_{\geq 0}(X,i(Y))\cong\mathrm{Mod}_R(H_0(X), Y)$. If you write out how maps on the left side look like, you will see that this essentially follows immediately (if you want details, say so and I will edit them in). Moreover, you will also see why it is necessary to restrict to $\mathrm{Chain}(R)_{\geq 0}$. The point is that if $X\in\mathrm{Chain}(R)$, then $X_0=\mathrm{ker}(d\colon X_0\to X_{-1})$ need not hold generally. Therefore, if I give you a map $H_0(X)\to Y$, you in principle only know how to map from $\mathrm{ker}(d\colon X_0\to X_{-1})$ to $Y$, and not necessarily from all of $X_0$ to $Y$. So you will not generally obtain a chain map $X\to i(Y)$. Note that any such chain map still gives you a map $H_0(X)\to Y$ because $H_0(i(Y))\cong Y$.
For an explicit counterexample, take $X$ to be the chain complex $$ \ldots\xrightarrow{\cdot 2} \mathbb{Z}/4\xrightarrow{\cdot 2}\mathbb{Z}/4\xrightarrow{\cdot 2}\mathbb{Z}/4\xrightarrow{\cdot 2}\ldots $$ and let $Y=\mathbb{Z}/4$. Then the only morphism $0\cong H_0(X)\to Y$ is the zero map. However, there is a chain map $X\to i(Y)$ with the map $\mathbb{Z}/4\xrightarrow{\cdot 2} Y$ in degree 0, and another one with the zero map $0\colon\mathbb{Z}/4\to Y$ in degree 0. Hence the adjunction fails in this case.
Edit: By the way, the functor $i\colon\mathrm{Mod}_R\to\mathrm{Chain}(R)_{\geq 0}$ is not only a right adjoint to $H_0$, but also itself a left adjoint to some other functor $G\colon\mathrm{Chain}(R)_{\geq 0}\to\mathrm{Mod}_R$. You may enjoy figuring out what this $G$ is.