The Stacks project

Proof. If $\mathcal{I}$ is the empty category, then the lemma is true. Thus we may assume $\mathcal{I}$ is nonempty. In this case we can write $\mathcal{I} = \coprod \mathcal{I}_ j$ where each $\mathcal{I}_ j$ is nonempty and satisfies the same property, see Lemma 4.19.6. Since $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} M = \coprod _ j \mathop{\mathrm{colim}}\nolimits _{\mathcal{I}_ j} M|_{\mathcal{I}_ j}$ this reduces the proof of (1) to the connected case.

Assume $\mathcal{I}$ is connected and $M \to N$ is injective, i.e., all the maps $M_ i \to N_ i$ are injective. We identify $M_ i$ with the image of $M_ i \to N_ i$, i.e., we will think of $M_ i$ as a subset of $N_ i$. We will use the description of the colimits given in Section 4.15 without further mention. Let $s, s' \in \mathop{\mathrm{colim}}\nolimits M_ i$ map to the same element of $\mathop{\mathrm{colim}}\nolimits N_ i$. Say $s$ comes from an element $m$ of $M_ i$ and $s'$ comes from an element $m'$ of $M_{i'}$. Then we can find a sequence $i = i_0, i_1, \ldots , i_ n = i'$ of objects of $\mathcal{I}$ and morphisms

and elements $n_{i_ j} \in N_{i_ j}$ mapping to each other under the maps $N_{i_{2k-1}} \to N_{i_{2k-2}}$ and $N_{i_{2k-1}} \to N_{i_{2k}}$ induced from the maps in $\mathcal{I}$ above with $n_{i_0} = m$ and $n_{i_{2n}} = m'$. We will prove by induction on $n$ that this implies $s = s'$. The base case $n = 0$ is trivial. Assume $n \geq 1$. Using the assumption on $\mathcal{I}$ we find a commutative diagram

We conclude that $m$ and $n_{i_2}$ map to the same element of $N_ w$ because both are the image of the element $n_{i_1}$. In particular, this element is an element $m'' \in M_ w$ which gives rise to the same element as $s$ in $\mathop{\mathrm{colim}}\nolimits M_ i$. Then we find the chain

and the elements $n_{i_ j}$ for $j \geq 3$ which has a smaller length than the chain we started with. This proves the induction step and the proof of (1) is complete.

Let $G$ be a group and let $\mathcal{I}$ be the one-object category with endomorphism monoid $G$. Then $\mathcal{I}$ satisfies the condition stated in the lemma because given $g_1, g_2 \in G$ we can find $h_1, h_2 \in G$ with $h_1 g_1 = h_2 g_2$. An diagram $M$ over $\mathcal{I}$ in $\textit{Ab}$ is the same thing as an abelian group $M$ with $G$-action and $\mathop{\mathrm{colim}}\nolimits _\mathcal {I} M$ is the coinvariants $M_ G$ of $M$. Take $G$ the group of order $2$ acting trivially on $M = \mathbf{Z}/2\mathbf{Z}$ mapping into the first summand of $N = \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2\mathbf{Z}$ where the nontrivial element of $G$ acts by $(x, y) \mapsto (x + y, y)$. Then $M_ G \to N_ G$ is zero. $\square$