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This is a simple question about Tor and Ext functors.

Let $R$ be a commutative ring, and let $0 \to M' \to M \to M'' \to 0$ and $0 \to N' \to N \to N'' \to 0$ be short exact sequences of $R$-modules. Then using the connecting homomorphisms from the long exact sequences, we can build the following diagram: \[ \begin{matrix} \operatorname{Tor}^R_i(M'',N'') & \longrightarrow & \operatorname{Tor}^R_{i-1}(M',N'') \\ \downarrow && \downarrow \\ \operatorname{Tor}^R_{i-1}(M'',N') & \longrightarrow & \operatorname{Tor}^R_{i-2}(M',N') \end{matrix} \] Is this diagram commutative? A similar question can be asked about the Ext functor.

I think the answer might depend on the definitions used: there are a couple of "equivalent" definitions for Tor/Ext. My motivation is actually to understand what equivalent should really mean here. (E.g. derived functors can be defined up to isomorphisms of $\delta$-functors, however $\operatorname{Tor}(\cdot, \cdot)$ seems to be something like a "2-variable $\delta$-functor".)

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1 Answer 1

For the Ext functor, you can get this using the Yoneda interpretation of $\operatorname{Ext}$. If $0 \to M'' \to L_* \to N' \to 0$ represents an element of $\operatorname{Ext}^n(M'',N')$, the image under the first connecting homomorphism is represented by the splice $$ 0 \to M' \to M \to L_* \to N' \to 0$$ and under the second $$ 0 \to M'' \to L_* \to N \to N'' \to 0$$ (MacLane's book on homology explains why the connecting homomorphism can be interpreted in this way). It's clear that you get the same sequence no matter which way round you do the two connecting homomorphisms.

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