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This question is regarding the Yoneda description of $Ext^{n}$ group of r modules M and N. I want to know that what is the inverse element of an n-extension of M by N. Let $X.= 0 \to N \to X_n \to X_{n-1}\to \ldots \to X_1 \to M \to 0$ be a representative of a class in $Ext^{n}$. I know the equivalence class of $I.=0 \to N \to N \to 0 \to \ldots \to M \to M \to 0$ is the zero element. Let $\overline{X}.= 0 \to N \to X_n \to X_{n-1}\to \ldots \to X_1 \to M \to 0$ with the map from $N \to X_n$ is the negative of the map from $N \to X_n$ in X. and all other maps are same as X.. I want to prove that Baer sum of X. and $\overline{X}.$ is equivalent to $I.$ Any idea of proof?

budi
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1 Answers1

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You can write the $n$-extension $X$ as the Yoneda composition $A\circ B$ of a $1$-extension $A$ and an $(n-1)$ extension $B$. Then $\overline X$ is the composition $\overline A\circ B$ of the $1$-extension $\overline A$ that you can imagine and the same $(n-1)$-extension $B$ as before. Since the equivalence relation for $n$-extensions is compatible with Yoneda composition and Baer sum (you should check this at some point), to show that the opposite of $A\circ B$ is equivalent to $\overline A\circ B$ it is enough to show that $A$ and $\overline A$ are opposite.

Do that. :-)

Mariano Suárez-Álvarez
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  • I am not going in the way of writing an n extension as a composition. i am directly defining an equivalence of n extensions using a commutative diagram that you can see here. http://www.uta.edu/honors/project/thesis/samples/Scie_Mathematics.pdf pages 26-27. In that way how can i prove this? – budi Sep 23 '16 at 03:49
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    Show that your equivalence is equivalent to the definition I want to use. – Mariano Suárez-Álvarez Sep 23 '16 at 03:50
  • With the notations as in the paper cited above can we get a map from T to N where T is the pushout used in the Baer sum to get a commutative diagram to show the equivalence? I stuck there. i am not able to get such a map. – budi Sep 23 '16 at 03:53