Talk:Ext functor

Left or right derived functors?

According to Osborne it is Left derived functors that are defined using projective resolutions and right derived functors that use injective resolutions. Doesn't this make Ext left derive.

IE start with a covariant functor Hom(X,-) and some object A, take a projective resolution of X, P_n --> ... --> P_0 --> X, then apply Hom, cut the end off: Hom(P_n,A) --> ... --> Hom(P_0,A), then take cohomology of the complex?

Fixed that. And a few things more. Michiexile 15:18, 9 February 2007 (UTC)[reply]

Isn't that Hom(P_n,A) <-- ... <-- Hom(P_0,A) ? Charles Matthews 15:53, 9 February 2007 (UTC)[reply]

It is. But in my rewrite, things are tweaked the right way around, regardless of what's in the discussion. Michiexile 21:51, 10 February 2007 (UTC)[reply]

Hom(A,-) is [covariant] left-exact, but contrary to what is in the article, Hom(-,B) is also left-exact. (See exact functor.) This is why both have right-derived functors. (Changed.) Tesseran 20:08, 11 February 2007 (UTC)[reply]

In answer to the original question: left-exact covariant functors, as well as right-exact contravariant functors, have their derived functors (right-derived and left-derived respectively) defined using injective resolutions. Right-exact covariant and left-exact contravariant functors have their derived functors (left-derived and right-derived resp.) defined using projective resolutions. Your statement is only true if all your original functors are covariant. Tesseran 20:12, 11 February 2007 (UTC)[reply]

Hom(-,B) is contravariant though. Which, upon a reread, meshes perfectly with what you wrote. Thanks for the correction. Michiexile 16:29, 12 February 2007 (UTC)[reply]

examples have dangling references

The examples have dangling references. Should R and M be the same? --MarSch 13:53, 2 May 2007 (UTC)[reply]

No, the first paragraph of the article introduces

{\mathcal {C}}

as the category of modules over some ring

R

. So, the statements in the example sequence just fix the category in which the example is exhibited, and

M

is the module of coefficients for the functor.

I'm not certain how to make this distinction clearer - suggestions are very welcome. Michiexile 22:33, 3 May 2007 (UTC)[reply]