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It is known that extensions $0 \to A \to B \to C \to 0$ are classified by $Ext^1(C, A)$. One can get such an element in two ways:

  • applying $RHom(\cdot, A)$, one gets $$0 \to Hom(C, A) \to Hom(B, A) \to Hom(A, A) \to Ext^1(C, A)$$ and takes the image of $id_A$ (as in Weibel's Homological algebra and two other books),

  • or applying $RHom(C, \cdot)$, one gets $$0 \to Hom(C, A) \to Hom(C, B) \to Hom(C, C) \to Ext^1(C, A)$$ and takes the image of $id_C$.

Is it true that the results are the same?

evgeny
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1 Answers1

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You can see that both ways give a bijection between the Yoneda $\operatorname{Ext}^1 (C,A)$ and the derived functor $R^1\operatorname{Hom} (-,A) (C)$ (resp. $R^1\operatorname{Hom} (C,-) (A)$). Weibel's book goes through the first bijection, but you can get the second bijection by literally dualizing all the steps of the proof (replacing projectives with injectives, etc.).

If you are asking about literally obtaining the same element, you should be careful: the two bijections are between the Yoneda Ext and $\operatorname{Hom} (-,-)$ derived in different arguments, and while the notation is "$\operatorname{Ext}^1 (C,A)$" in both cases, these two are different sets.

  • It does make sense to ask whether you get the same element: while the Ext group is defined only up to isomorphism, the isomorphism is canonical. – Eric Wofsey Jan 04 '17 at 23:19
  • Here we need to somehow identify $\operatorname{Hom} (-,-)$ derived in different arguments. Actually, one way to do that is to use the two isomorphisms with the Yoneda Ext. –  Jan 05 '17 at 01:07
  • Oh, good point. There are various canonical ways to identify them but I'm not sure what the "right" one is (I don't immediately see a way to do it in terms of any universal property), nor is it obvious that all of these canonical ways are actually the same. – Eric Wofsey Jan 05 '17 at 02:34