The Stacks project

Proof. We will prove this in the case $S$ is a left multiplicative system. The case where $S$ is a right multiplicative system is dual. Suppose that $X, Y$ are objects of $\mathcal{C}$ and that $\alpha , \beta : X \to Y$ are morphisms in $S^{-1}\mathcal{C}$. According to Categories, Lemma 4.27.5 we may represent these by pairs $s^{-1}f, s^{-1}g$ with common denominator $s$. Suppose that we have a pre-additive structure on $S^{-1}\mathcal{C}$ and that $Q$ is additive. Then

Hence, the pre-additive structure on $S^{-1}\mathcal{C}$ is unique if it exists. Correspondingly, we define $\alpha + \beta $ to be the equivalence class of $s^{-1}(f + g)$ and in the rest of the proof we show that this is well defined and that composition is bilinear. Once this is done it is clear that $Q$ is an additive functor.

Let us show construction above is well defined. An abstract way of saying this is that filtered colimits of abelian groups agree with filtered colimits of sets and to use Categories, Equation (4.27.7.1). We can work this out in a bit more detail as follows. Say $s : Y \to Y_1$ and $f, g : X \to Y_1$. Suppose we have a second representation of $\alpha , \beta $ as $(s')^{-1}f', (s')^{-1}g'$ with $s' : Y \to Y_2$ and $f', g' : X \to Y_2$. By Categories, Remark 4.27.7 we can find a morphism $s_3 : Y \to Y_3$ and morphisms $a_1 : Y_1 \to Y_3$, $a_2 : Y_2 \to Y_3$ such that $a_1 \circ s = s_3 = a_2 \circ s'$ and also $a_1 \circ f = a_2 \circ f'$ and $a_1 \circ g = a_2 \circ g'$. Hence we see that $s^{-1}(f + g)$ is equivalent to

which is equivalent to $(s')^{-1}(f' + g')$.

Fix $s : Y \to Y'$ and $f, g : X \to Y'$ with $\alpha = s^{-1}f$ and $\beta = s^{-1}g$ as morphisms $X \to Y$ in $S^{-1}\mathcal{C}$. To show that composition is bilinear first consider the case of a morphism $\gamma : Y \to Z$ in $S^{-1}\mathcal{C}$. Say $\gamma = t^{-1}h$ for some $h : Y \to Z'$ and $t : Z \to Z'$ in $S$. Using LMS2 we choose morphisms $a : Y' \to Z''$ and $t' : Z' \to Z''$ in $S$ such that $a \circ s = t' \circ h$. Picture

Then $\gamma \circ \alpha = (t' \circ t)^{-1}(a \circ f)$ and $\gamma \circ \beta = (t' \circ t)^{-1}(a \circ g)$. Hence we see that $\gamma \circ (\alpha + \beta )$ is represented by $(t' \circ t)^{-1}(a \circ (f + g)) = (t' \circ t)^{-1}(a \circ f + a \circ g)$ which represents $\gamma \circ \alpha + \gamma \circ \beta $.

Finally, assume that $\delta : W \to X$ is another morphism of $S^{-1}\mathcal{C}$. Say $\delta = r^{-1}i$ for some $i : W \to X'$ and $r : X \to X'$ in $S$. We claim that we can find a morphism $s' : Y' \to Y''$ in $S$ and morphisms $a'', b'' : X' \to Y''$ such that the following diagram commutes

Namely, using LMS2 we can first choose $s_1 : Y' \to Y_1$, $s_2 : Y' \to Y_2$ in $S$ and $a : X' \to Y_1$, $b : X' \to Y_2$ such that $a \circ r = s_1 \circ f$ and $b \circ r = s_2 \circ f$. Then using that the category $Y'/S$ is filtered (see Categories, Remark 4.27.7), we can find a $s' : Y' \to Y''$ and morphisms $a' : Y_1 \to Y''$, $b' : Y_2 \to Y''$ such that $s' = a' \circ s_1$ and $s' = b' \circ s_2$. Setting $a'' = a' \circ a$ and $b'' = b' \circ b$ works. At this point we see that the compositions $\alpha \circ \delta $ and $\beta \circ \delta $ are represented by $(s' \circ s)^{-1}(a'' \circ i)$ and $(s' \circ s)^{-1}(b'' \circ i)$. Hence $\alpha \circ \delta + \beta \circ \delta $ is represented by $(s' \circ s)^{-1}(a'' \circ i + b'' \circ i) = (s' \circ s)^{-1}((a'' + b'') \circ i)$ which by the diagram again is a representative of $(\alpha + \beta ) \circ \delta $. $\square$

Proof. By Lemma 12.8.1 we see that $S^{-1}\mathcal{C}$ is preadditive and that $Q$ is additive. Recall that the functor $Q$ commutes with finite colimits (resp. finite limits), see Categories, Lemmas 4.27.9 and 4.27.17. We conclude that $S^{-1}\mathcal{C}$ has a zero object and direct sums, see Lemmas 12.3.2 and 12.3.4. $\square$

Proof. If (2) holds we see that $0 = Q(0) : Q(X) \to Q(Y)$ is an isomorphism. In the additive category $S^{-1}\mathcal{C}$ this implies that $Q(X) = 0$. Hence (2) $\Rightarrow $ (1). Similarly, (3) $\Rightarrow $ (1). Suppose that $Q(X) = 0$. This implies that the morphism $f : 0 \to X$ is transformed into an isomorphism in $S^{-1}\mathcal{C}$. Hence by Categories, Lemma 4.27.21 there exists a morphism $g : Z \to 0$ such that $fg \in S$. This proves (1) $\Rightarrow $ (3). Similarly, (1) $\Rightarrow $ (2). $\square$

Proof. Assume $S$ is a left multiplicative system. Let $a : X \to Y$ be a morphism of $S^{-1}\mathcal{A}$. Then $a = s^{-1}f$ for some $s : Y \to Y'$ in $S$ and $f : X \to Y'$. Since $Q(s)$ is an isomorphism we see that the existence of $\mathop{\mathrm{Coker}}(a : X \to Y)$ is equivalent to the existence of $\mathop{\mathrm{Coker}}(Q(f) : X \to Y')$. Since $\mathop{\mathrm{Coker}}(Q(f))$ is the coequalizer of $0$ and $Q(f)$ we see that $\mathop{\mathrm{Coker}}(Q(f))$ is represented by $Q(\mathop{\mathrm{Coker}}(f))$ by Categories, Lemma 4.27.9. This proves (1).

Part (2) is dual to part (1).

If $S$ is a multiplicative system, then $S$ is both a left and a right multiplicative system. Thus we see that $S^{-1}\mathcal{A}$ has kernels and cokernels and $Q$ commutes with kernels and cokernels. To finish the proof of (3) we have to show that $\mathop{\mathrm{Coim}}= \mathop{\mathrm{Im}}$ in $S^{-1}\mathcal{A}$. Again using that any arrow in $S^{-1}\mathcal{A}$ is isomorphic to an arrow $Q(f)$ we see that the result follows from the result for $\mathcal{A}$. $\square$