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Within the book An introduction to homological algebra by Weibel, I am trying to prove the following isomorphism, but I am not sure this is true. But I really want to know how to prove or disprove it.

Notation: Let $\mathfrak{g}$ be a finite dimensional Lie algebra over a field. Let $\text{H}^i_{\text{Lie}}(\mathfrak{g},A)$ be the i-th cohomology group of $\mathfrak{g} $ with the coefficients in the $\mathfrak{g}$-module $A$.

Page 226: Let $M,N$ be $\mathfrak{g}$-modules. Then we have the following natural isomorphism of $\delta$ functors:

$$ \text{Ext}^\ast_{U \mathfrak{g}}(M,N) \cong \text{H}^\ast_{\text{Lie}}(\mathfrak{g}, \text{Hom}_{k}(M,N)) $$

where $\ast$ is a non-negative integer.

Thanks very much!

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closed as off-topic by Fernando Muro, Benjamin Steinberg, José Figueroa-O'Farrill, Andrés E. Caicedo, Angelo Jul 2 '13 at 17:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Benjamin Steinberg, AndrĂ©s E. Caicedo, Angelo
If this question can be reworded to fit the rules in the help center, please edit the question.

This isn't really a research-level question. You are quoting Exercise 7.3.5 in Weibel's concise Chapter 7 on Lie algebra cohomology; it deals with identification of derived functors based on Exercise 7.3.4. This is closely parallel to his earlier treatment of groups, so Weibel is leaving most details to the reader. You have to assimilate his earlier chapters well in order to follow Chapter 7. – Jim Humphreys Jul 2 '13 at 15:39
One way to do this is to see both sides of the isomrphism as functors of $N$, and prove they are $\partial$-functors: this reduces the problem to degree zero, where a direct computation does the trick. – Mariano Suárez-Alvarez Jul 3 '13 at 0:58