Category of fractions: transitivity and cancellation property 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.
 A: 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.
