It is the category of functors from $\mathbb N_{\geq }$ to abelian groups. Are the morphisms natural transformations?
I am learning basic homology theory but am not quite sure if I have understood it correctly.
Suppose $(G_n,\psi_n:G_{n+1}\to G_{n})\in \mathbf{Fun}(\mathbb N,\text{Ab})$ is an injective object, so that for any monomorphism $T:(H_n,\tau_n)\to (K_n,\sigma_n)$ and morphism $S:(H_n,\tau_n)\to (G_n,\psi_n)$, there is $R:(K_n,\sigma_n)\to (G_n,\psi_n)$ such that $RT=S$.
Suppose $T:(H_n,\tau_n)\to (K_n,\sigma_n)$ is monic, then for any different $U,V:(J_n,\eta_n)\to (H_n,\tau_n)$, $TU\neq TV$. Let $J_{n_0}=\ker T_{n_0}$ and $J_n=0$ for others. Let $U_{n_0}:J_{n_0}\to H_{n_0}$ be the inclusion map and $V=0$. Since $TU=TV=0$, $U=V$ and therefore $\ker T_{n_0}$ must be $0$. This shows $T$ is monic iff each $T_n$ is monic.
Next suppose $G_n$ are divisible groups, then $(G_n,\psi_n)$ is injective, because, for any $(H_n,\tau_n)\leq (K_n,\sigma_n)$ ,take $x\in K_n\backslash H_n$, construct $H_n'=\langle H_n,x\rangle$ and define $$S'_n(x)=\left\{\begin{array}{}S_n(kx)/k&,kx\in H_n\\0&,others \end{array}\right.$$, then $S':(H',\tau'_n)\to (G_n,\psi_n)$ extends $S$. Eventually, using Zorn's lemma, $S$ is extended to some $R$.
Now I want to find a injective resolution for $(G_n,\psi_n)$. Let $G^1_n$ be the divisible group generated by $G_n$ and let $H^1_n=G_n^1/G_n$, let $G_n^{k+1}$ be the divisible group generated by $H_n^k$ and let $H_n^{k+1}=G_n^{k+1}/H_n^k$. Then $\mathbf G\to \mathbf G^1\to \mathbf G^2\to ...$ is an injective resolution for $\mathbf G=(G_n,\psi_n)$.
As an example, take $G_n=\oplus_{k=1}^n\mathbb Z/2\mathbb Z$ and let $\psi_n$ maps $(a_1,...,a_{n+1})$ to $(a_1,...,a_n)$. Now $G_n^1$ would be $\oplus_{k=1}^n \mathbb Q/\mathbb Z$ and $d^1_n:G_n\to G_n^1$ maps $(0,...,1,...,0)$ to $(0,...,[1/2],...,0)$. Therefore, $H_n^1=\oplus_{k=1}^n \mathbb Q/\mathbb Z/\{0,[1/2]\}$, however, $\simeq \oplus_{k=1}^n \mathbb Q/\mathbb Z$. Since $H_n^1$ is injective itself, $G_n^2=H_n^1$ and $G_n^3=H_n^2=0$.
Now I want to apply $\lim_\leftarrow$ and its right derived functor to $G_n$. It is obvious that $\lim_\leftarrow G_n=\oplus_{k\in\mathbb N}\mathbb Z/2\mathbb Z$ (with homomorphisms $\alpha_n$ mapping $(a_1,a_2,...)$ to $(a_1,...,a_n)$), since if $A$ is an abelian group and $\beta_n$ maps $x\in A$ to $(a_1,...,a_n)\in G_n$, then by defining $\gamma(x)=(a_1,a_2,...)$, $\gamma:A\to \oplus_{k\in\mathbb N}\mathbb Z/2\mathbb Z$ makes $\alpha_n\gamma=\beta_n$.
To calculate $\lim_\leftarrow^1 G_n$, let $G=\oplus_{k\in\mathbb N}\mathbb Z/2\mathbb Z$, $G^1=G^2=\oplus_{k\in \mathbb N} \mathbb Q/\mathbb Z$, $G^3=0$. Denote by $\theta^2$ the morphism from $G^1\to G^2$ induced by $\lim_\leftarrow$ and $d_n^2:G_n^1\to H_n^1\to G_n^2;x\mapsto 2x$, then $\lim^0_\leftarrow G_n=\ker \theta^2=\oplus_{k\in\mathbb N}\mathbb Z/2\mathbb Z$ and $\lim^1_\leftarrow G_n=\ker \theta^3/\text{Im } \theta^2=\text{coker }\theta^2=\oplus_{k\in \mathbb N} \mathbb Q/\mathbb Z$.
Does this seem correct?