Lemma 4.27.18. Let $\mathcal{C}$ be a category. Let $S$ be a right multiplicative system. If $f : X \to Y$, $f' : X' \to Y'$ are two morphisms of $\mathcal{C}$ and if

$\xymatrix{ Q(X) \ar[d]_{Q(f)} \ar[r]_ a & Q(X') \ar[d]^{Q(f')} \\ Q(Y) \ar[r]^ b & Q(Y') }$

is a commutative diagram in $S^{-1}\mathcal{C}$, then there exist a morphism $f'' : X'' \to Y''$ in $\mathcal{C}$ and a commutative diagram

$\xymatrix{ X \ar[d]_ f & X'' \ar[l]^ s \ar[d]^{f''} \ar[r]_ g & X' \ar[d]^{f'} \\ Y & Y'' \ar[l]_ t \ar[r]^ h & Y' }$

in $\mathcal{C}$ with $s, t \in S$ and $a = gs^{-1}$, $b = ht^{-1}$.

Proof. This lemma is dual to Lemma 4.27.10. $\square$

Comment #1585 by Darij Grinberg on

I have changed the proof of Lemma {lemma-left-localization-diagram} in one of my last commits, but not that of Lemma {lemma-right-localization-diagram}. I guess the same changes should be done here (in particular, I don't believe that we can choose $d : X''' \to X''$ to lie in $S$, but I don't believe that we need this either). But I am wondering whether it is really worth doing this proof, given that the lemma is just the dual of Lemma {lemma-left-localization-diagram}. (Or so it looks to me -- I have not compared them thoroughly.)

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