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.

  • 1
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.