Lemma 4.27.21. Let $\mathcal{C}$ be a category and let $S$ be a multiplicative system. Denote $Q : \mathcal{C} \to S^{-1}\mathcal{C}$ the localization functor. The set

\[ \hat S = \{ f \in \text{Arrows}(\mathcal{C}) \mid Q(f) \text{ is an isomorphism}\} \]

is equal to

\[ S' = \{ f \in \text{Arrows}(\mathcal{C}) \mid \text{there exist }g, h\text{ such that }gf, fh \in S\} \]

and is the smallest saturated multiplicative system containing $S$. In particular, if $S$ is saturated, then $\hat S = S$.

**Proof.**
It is clear that $S \subset S' \subset \hat S$ because elements of $S'$ map to morphisms in $S^{-1}\mathcal{C}$ which have both left and right inverses. Note that $S'$ satisfies MS4, and that $\hat S$ satisfies MS1. Next, we prove that $S' = \hat S$.

Let $f \in \hat S$. Let $s^{-1}g = ht^{-1}$ be the inverse morphism in $S^{-1}\mathcal{C}$. (We may use both left fractions and right fractions to describe morphisms in $S^{-1}\mathcal{C}$, see Lemma 4.27.19.) The relation $\text{id}_ X = s^{-1}gf$ in $S^{-1}\mathcal{C}$ means there exists a commutative diagram

\[ \xymatrix{ & X' \ar[d]^ u & \\ X \ar[ru]^{gf} \ar[r]^{f'} \ar[rd]_{\text{id}_ X} & X'' & X \ar[lu]_ s \ar[l]_{s'} \ar[ld]^{\text{id}_ X} \\ & X \ar[u]_ v & } \]

for some morphisms $f', u, v$ and $s' \in S$. Hence $ugf = s' \in S$. Similarly, using that $\text{id}_ Y = fht^{-1}$ one proves that $fhw \in S$ for some $w$. We conclude that $f \in S'$. Thus $S' = \hat S$. Provided we prove that $S' = \hat S$ is a multiplicative system it is now clear that this implies that $S' = \hat S$ is the smallest saturated system containing $S$.

Our remarks above take care of MS1 and MS4, so to finish the proof of the lemma we have to show that LMS2, RMS2, LMS3, RMS3 hold for $\hat S$. Let us check that LMS2 holds for $\hat S$. Suppose we have a solid diagram

\[ \xymatrix{ X \ar[d]_ t \ar[r]_ g & Y \ar@{..>}[d]^ s \\ Z \ar@{..>}[r]^ f & W } \]

with $t \in \hat S$. Pick a morphism $a : Z \to Z'$ such that $at \in S$. Then we can use LMS2 for $S$ to find a commutative diagram

\[ \xymatrix{ X \ar[d]_ t \ar[r]_ g & Y \ar[dd]^ s \\ Z \ar[d]_ a \\ Z' \ar[r]^{f'} & W } \]

and setting $f = f' \circ a$ we win. The proof of RMS2 is dual to this. Finally, suppose given a pair of morphisms $f, g : X \to Y$ and $t \in \hat S$ with target $X$ such that $ft = gt$. Then we pick a morphism $b$ such that $tb \in S$. Then $ftb = gtb$ which implies by LMS3 for $S$ that there exists an $s \in S$ with source $Y$ such that $sf = sg$ as desired. The proof of RMS3 is dual to this.
$\square$

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