Founding of homological without quite involving derived categories - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T12:37:22Zhttp://mathoverflow.net/feeds/question/10997http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/10997/founding-of-homological-without-quite-involving-derived-categoriesFounding of homological without quite involving derived categoriesAllen Knutson2010-01-07T01:49:21Z2010-01-07T19:28:22Z
<p>I am looking at the foundations of homological algebra, e.g. the introduction
of Ext and Tor, and am unsatisfied. The references I look at start with
"this is called a projective module, this is called a projective resolution,
now pick one and use it to define the right derived functors of your
left exact functor". I would like to see a presentation more along the
following lines:</p>
<ol>
<li><p>The functor Hom(A,*), applied to a short exact sequence of modules,
doesn't produce another such. An oracle tells us that it does produce
a long exact sequence; what could it be?</p></li>
<li><p>We already know (from antiquity) that a short exact sequence of
<em>complexes</em> induces a long exact sequence on cohomology.</p></li>
<li><p>But in #1 we put in modules, not complexes. So let's fix that by hoping
that Hom(A,*) can be extended in a natural way to the category of complexes
(and really, to descend to the derived category).</p></li>
<li><p>Such an extension might be required to have the following properties: ???</p></li>
<li><p>Now I'd like it to be easy to see that the extension is unique if it
exists. When is it easy to compute? At this point I'd like the definition
of "projective module" to suggest itself.</p></li>
<li><p>Finally, the usual boring checks that using projective resolutions
to define it, the extension does indeed exist.</p></li>
</ol>
<p>One way to answer this is to say "In part 4, define the derived category,
and its t-structure, then ask that the extension be exact in the
appropriate sense". I'm hoping to avoid going quite <em>that</em> far, or at least,
doing it in a way that doesn't involve introducing too many more definitions.</p>
http://mathoverflow.net/questions/10997/founding-of-homological-without-quite-involving-derived-categories/11063#11063Answer by David Speyer for Founding of homological without quite involving derived categoriesDavid Speyer2010-01-07T19:28:22Z2010-01-07T19:28:22Z<p>On rereading your question, Jacobson is not what you want after all.</p>
<p>Anton gave a very nice answer along these lines <a href="http://mathoverflow.net/questions/1151/sheaf-cohomology-and-injective-resolutions" rel="nofollow">here</a>. In comments to that post, Tyler Lawson recommends Cartan and Eilenberg.</p>