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Given a category $\mathcal{C}$, and a right calculus of fractions $\Sigma$. We can construct the category of fractions $\mathcal{C}[\Sigma^{-1}]$ which has the same objects as $\mathcal{C}$, and homsets given by $$ \mathcal{C}[\Sigma^{-1}](x,y) = \{y\xleftarrow{\gamma}u\xrightarrow{\sigma}x|\sigma\in\Sigma\}/\mathord{\sim} $$ where $y\xleftarrow{\gamma}u\xrightarrow{\sigma}x\sim y\xleftarrow{\gamma'}u\xrightarrow{\sigma'}x$ iff there is an object $v$ and morphisms $\tau\colon v\to u$, $\tau'\colon v\to u'$ in $\Sigma$, such that $\gamma\circ\tau = \gamma'\circ\tau'$ and $\sigma\circ\tau=\sigma'\circ\tau$. It's obvious that this relation is symmetric and reflexive, but I don't manage to show transitivity. I.e. given $$ y\xleftarrow{\gamma_1}u_1\xrightarrow{\sigma_1}x\sim y\xleftarrow{\gamma_2}u_2\xrightarrow{\sigma_2}x\sim y\xleftarrow{\gamma_3}u_3\xrightarrow{\sigma_3}x $$ I get a diagram $$ \matrix{ &&y&=&y&=&y\\ &\nearrow&&&\uparrow&&&\nwarrow\\ u_1&\xleftarrow{\tau_1}&v&\xrightarrow{\tau_2}&u_2&\xleftarrow{\tau_2'}&v'\xrightarrow{\tau_3}&u_3\\ &\searrow&&&\downarrow&&&\swarrow\\ &&x&=&x&=&x\\} $$ where the upwards arrows are the respective $\gamma_i$ and the downward arrows are the respective $\sigma_i$. But I'm stuck from there on.

Moreover, when defining a right calculus on fractions $\Sigma$, we impose a right cancellability condition, i.e. given an arrow $\sigma\colon y\to z$ in $\Sigma$ and a pair of parallel morphisms $f,g\colon x→y$ such that $\sigma\circ f=\sigma\circ g$, there is an arrow $\sigma′\colon w→x$ in $\Sigma$ such that $f\circ\sigma′=g\circ\sigma′$. Provided we don't need this to proof transitivity, I don't see why it is necessary, since it doesn't seem to be required for any other part of the proof that $\mathcal{C}[\Sigma^{-1}]$ is a well defined category.

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    $\begingroup$ Have you already had a look at the textbook by Gelfand & Manin on homological algebra? There is a proof of transitivity for the eq. relation of morphisms in derived categories (it applies to your case) $\endgroup$ – Avitus Sep 25 '13 at 11:02
  • $\begingroup$ Oh, thanks for the hint. I've actually have a copy of that in my bureau, gonna check that out tomorrow. $\endgroup$ – roman Sep 25 '13 at 13:46
  • $\begingroup$ you are welcome... pag. 149-150 :-) $\endgroup$ – Avitus Sep 25 '13 at 15:12
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Ok, following the hint by user Avitus, I found the answer in Gelfand & Manin Methods of Homological Algebra:

First, given the morphisms $\sigma_1\circ\tau_1$ and $\sigma_2\circ\tau_2'$, we can use the right extension condition to find an object $w$, together with morhisms $k\in\Sigma(w,v)$, $k'\in\mathcal{C}(w,v')$, such that $$\sigma_1\circ\tau_1\circ k = \sigma_2\circ\tau_2'\circ k'\text{.}$$ Moreover, $$ \sigma_1\circ\tau_1\circ k = \sigma_2\circ\tau_2\circ k\text{.}$$ Hence $\sigma_2$ equalizes $\tau_2'\circ k'$ and $\tau_2\circ k$. Thus, using the right cancellability condition, we can find an object $w'$, and a morphism $\kappa\in\Sigma(w',w)$, such that $\tau_2'\circ k'\circ\kappa= \tau_2\circ k\circ\kappa$. We observe that $$ \sigma_1\circ\tau_1\circ k\circ\kappa\\ =\sigma_2\circ\tau_2\circ k\circ\kappa\\ =\sigma_2\circ\tau_2'\circ k'\circ\kappa\\ =\sigma_3\circ\tau_3\circ k'\circ\kappa $$ and doing the same calculations with $\gamma_i$ instead of $\sigma_i$ we get $$ \gamma_1\circ\tau_1\circ k\circ\kappa=\gamma_3\circ\tau_3\circ k'\circ\kappa\text{.} $$ Hence the diagramm $$ \matrix{ &&y\\ &\nearrow&&\nwarrow\\ u_1&\xleftarrow{\tau_1\circ k\circ\kappa}&w'&\xrightarrow{\tau_3\circ k'\circ\kappa}&u_3\\ &\searrow&&\swarrow\\ &&x\\} $$ commutes, which yields $y\xleftarrow{\gamma_1}u_1\xrightarrow{\sigma_1}x\sim y\xleftarrow{\gamma_3}u_3\xrightarrow{\sigma_3}x$ and completes the proof.

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