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I'm learning Algebraic Topology by Massey . In chapter $6$ Homology with Arbitrary Coefficient group , he assumes knowledge of The Tor functor . So I tried to find the definition of Tor on Internet , specifically in Gradute texts - Rotman .

For each abelian group $A$ , choose an exact sequence $0 \to R \to F \to A \to 0$ with $F$ is free abelian . For any abelian group $B$ , define

$$\text{Tor}(A,B) = ker( (R \to F) \otimes 1_{B} )$$

He also wrote : " If we delete $A$ then $C_{*} = 0 \to R \to F \to 0$ is a chain complex as is $C_{*} \otimes B = 0 \to R \otimes B \to F \otimes B \to 0$ ( just attach a sequence of zeros ) . Hence $\text{Tor}(A,B) = H_{1}(C_{*} \otimes B)$ "

I don't understand why has this definition ( I knew there is another definition with show all relations and generators ) and the last paragraph ?

Finally for a short exact sequence of abelian groups I want to prove the following ( here I understand Tor in another way which is deffirent from above )

$$0 \to A' \to A \to A'' \to 0$$

We also have the exact sequence :

$$0 \to \text{Tor}(A',B) \to \text{Tor}(A,B) \to \text{Tor}(A'',B) \to A' \otimes B \to A \otimes B \to A'' \otimes B \to 0$$

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It's not clear to which alternative definition of $\operatorname{Tor}$ you refer to.

Assuming you are familiar with some properties of tensor products, here's a more or less self-contained proof, which may be extracted from any textbook on homological algebra.

  • The definition you give is correct: to calculate $\operatorname{Tor} (A,B)$, one should pick a short exact sequence $$\tag{1} 0 \to F_1 \to F_0 \to A \to 0$$ where $F_0$ is a free abelian group (and hence $F_1$ is free as well, being a subgroup of $F_0$). Then you should tensor it with $B$, and the result is $$0\to \operatorname{Tor} (A,B) \to F_1\otimes B \to F_0\otimes B\to A\otimes B\to 0$$ In other words, we declare $\operatorname{Tor} (A,B)$ to be the kernel of $F_1\to F_0$ tensored with $B$. (Note that $F_0\otimes B\to A\otimes B$ will still be an epimorphism. It is the monomorphism $F_1 \to F_0$ which after tensoring with $B$ may have a nonrivial kernel.)

(One should also justify why different choices of (1) lead to the same $\operatorname{Tor} (A,B)$. This is true, but let me omit this verification, as this post is already long enough.)

  • Note that if you have a short exact sequence of free abelian groups $$0 \to F' \to F \to F'' \to 0$$ then tensoring with $B$ gives you again a short exact sequence $$0 \to F'\otimes B \to F\otimes B \to F''\otimes B \to 0$$ This agrees with our definition according to which $\operatorname{Tor} (A,B) = 0$ if $A$ is free.

  • If you erase $A$ from the short exact sequence (1), you obtain a chain complex $$F_\bullet\colon\quad 0 \to F_1 \to F_0 \to 0$$ where $F_1$ sits in degree $1$ and $F_0$ sits in degree $0$. If you tensor this chain complex with $B$, you get a chain complex $$F_\bullet\otimes B\colon\quad 0 \to F_1\otimes B \to F_0\otimes B \to 0$$ Its homology in degree $0$ is given by $$H_0 (F_\bullet\otimes B) = \frac{\ker (F_0\otimes B \to 0)}{\operatorname{im} (F_1\otimes B \to F_0\otimes B)} = \operatorname{coker} (F_1\otimes B \to F_0\otimes B) \cong A\otimes B,$$ and its homology in degree $1$ is given by $$H_1 (F_\bullet\otimes B) = \frac{\ker (F_1\otimes B \to F_0\otimes B)}{\operatorname{im} (0 \to F_1\otimes B)} = \ker (F_1\otimes B \to F_0\otimes B),$$ which is $\operatorname{Tor} (A,B)$ by our definition.

  • The point of the above is the following. If you have a short exact sequence of abelian groups $$\tag{2} 0 \to A' \to A \to A'' \to 0$$ then to calculate $\operatorname{Tor} (A',B)$, $\operatorname{Tor} (A,B)$, $\operatorname{Tor} (A'',B)$, you should pick short exact sequences $$\tag{3} 0 \to F_1' \to F_0' \to A' \to 0$$ $$\tag{4} 0 \to F_1 \to F_0 \to A \to 0$$ $$\tag{5} 0 \to F_1'' \to F_0'' \to A'' \to 0$$ In fact you may do even better: first pick (3) and (5):

$$\require{AMScd}\begin{CD} @.0 @.@. 0\\ @.@VVV @. @VVV \\ @.F_1' @.@. F_1''\\ @.@VVV @. @VVV \\ @.F_0' @.@. F_0'' \\ @.@VVV @. @VVV \\ 0 @>>> A' @>>> A @>>> A'' @>>> 0 \\ @.@VVV @. @VVV \\ @.0 @.@. 0 \end{CD}$$ and then it is possible to build a short exact sequence (4) that gives a commutative diagram with exact rows and columns

$$\require{AMScd}\begin{CD} @.0 @. 0 @. 0\\ @.@VVV @VVV @VVV \\ 0 @>>> F_1' @>>> F_1 @>>> F_1'' @>>> 0 \\ @.@VVV @VVV @VVV \\ 0 @>>> F_0' @>>> F_0 @>>> F_0'' @>>> 0 \\ @.@VVV @VVV @VVV \\ 0 @>>> A' @>>> A @>>> A'' @>>> 0 \\ @.@VVV @VVV @VVV \\ @.0 @. 0 @. 0 \end{CD}$$

(This is known as the "horseshoe lemma", and here's how to do that: you may take $F_0 = F_0'\oplus F_0''$. There is a homomorphism $F_0'\to A$, which is just the composition of arrows $F_0'\to A$ and $A'\to A$, and also a homomorphism $F_0''\to A$, which is a lifting of the homomorphism $F_0'' \to A$ along the epimorphism $A\to A''$. Such a lifting is possible since $F_0''$ is free. These two arrows give you an arrow $F_0'\oplus F_0'' \to A$. Now repeat the same to obtain $F_1 = F_1'\oplus F_1''$ together with an arrow $F_1\to F_0$.)

So starting from (4), we obtained a short exact sequence of chain complexes $$\tag{6} 0 \to F_\bullet' \to F_\bullet \to F_\bullet'' \to 0$$ Now as these complexes consist of free abelian groups, after tensoring with $B$, we still get a short exact sequence of complexes $$0 \to F_\bullet'\otimes B \to F_\bullet\otimes B \to F_\bullet''\otimes B \to 0$$ It remains to invoke the following important theorem:

A short exact sequence of complexes $$0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0$$ induces a long exact sequence of homology groups $$\cdots \to H_n (A_\bullet) \to H_n (B_\bullet) \to H_n (C_\bullet) \to H_{n-1} (A_\bullet) \to \cdots$$

In our case this long exact sequence reads

\begin{multline} 0 \to H_1 (F_\bullet'\otimes B) \to H_1 (F_\bullet\otimes B) \to H_1 (F_\bullet''\otimes B) \to\\ H_0 (F_\bullet'\otimes B) \to H_0 (F_\bullet\otimes B) \to H_0 (F_\bullet''\otimes B) \to 0 \end{multline}

($H_2 = 0$ in our case). This is precisely the exact sequence with $\operatorname{Tor}$s!

  • There is no way you can go through an algebraic topology class without seeing the important theorem that I quote above. It is proved using the snake lemma, but in our case everything is even easier. The short exact sequence of complexes (6) is nothing more than the following commutative diagram with exact rows

$$\require{AMScd}\begin{CD} 0 @>>> F_1' @>>> F_1 @>>> F_1'' @>>> 0 \\ @.@VVV @VVV @VVV \\ 0 @>>> F_0' @>>> F_0 @>>> F_0'' @>>> 0 \end{CD}$$

After tensoring this with $B$, we obtain a diagram (still with exact rows, because the involved groups are free)

$$\require{AMScd}\begin{CD} 0 @>>> F_1'\otimes B @>>> F_1\otimes B @>>> F_1''\otimes B @>>> 0 \\ @.@VVV @VVV @VVV \\ 0 @>>> F_0'\otimes B @>>> F_0\otimes B @>>> F_0''\otimes B @>>> 0 \end{CD}$$

Applying the snake lemma to this diagram, you get

\begin{multline} 0 \to \ker (F_1'\otimes B\to F_0'\otimes B) \to \ker (F_1\otimes B\to F_0\otimes B) \to \ker (F_1''\otimes B\to F_0''\otimes B) \to \\ \operatorname{coker} (F_1'\otimes B\to F_0'\otimes B) \to \operatorname{coker} (F_1\otimes B\to F_0\otimes B) \to \operatorname{coker} (F_1''\otimes B\to F_0''\otimes B) \to 0 \end{multline}

which is again the exact sequence we are looking for.

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  • $\begingroup$ Oh , I knew these lemmas but didn't know their name . $\endgroup$ – Gankedbymom Apr 26 '17 at 17:07

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