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long exact sequence in cohomology
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(Theorem)
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Theorem 1 A morphism of cochain complexes $f:\mathcal{A} \to \mathcal{B}$ induces group homomorphisms $H^n(\mathcal{A}) \to H^n(\mathcal{B})$ for every $n$ .
Proof. This follows trivially from the fact that images and kernels of the cochain maps of $\mathcal{A}$ are mapped under $f$ to the images and kernels of the cochain maps of $\mathcal{B}$ . 
Definition 2 If $\alpha : \mathcal{A}\to\mathcal{B}$ and $\beta : \mathcal{B}\to\mathcal{C}$ are morphisms of cochain complexes, we say that $$ 0 \to \mathcal{A} \xrightarrow{\alpha} \mathcal{B} \xrightarrow{\beta} \mathcal{C} \to 0 $$ is a short exact sequence of cochain complexes if for every $n$ , $$ 0 \to A^n \xrightarrow{\alpha_n} B^n \xrightarrow{\beta_n} C^n \to 0 $$ is short exact.
Theorem 2 Let $$ 0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0 $$ be a short exact sequence of cochain complexes with $A^n = B^n = C^n = 0$ for $n<0$ . Then there is a long exact sequence of cohomology groups $$ 0 \to H^0(\mathcal{A}) \to H^0(\mathcal{B}) \to H^0(\mathcal{C}) \to H^1(\mathcal{A}) \to H^1(\mathcal{B}) \to H^1(\mathcal{C}) \to H^2(\mathcal{A}) \to \cdots $$
The proof that for each $n$ , $H^n(\mathcal{A}) \to H^n(\mathcal{B}) \to H^n(\mathcal{C})$ is exact is straightforward from the above. The interesting part of the proof is in defining the connecting homomorphism $$ \delta_n : H^n(\mathcal{C}) \to H^{n+1}(\mathcal{A}) $$ To define $\delta_n$ , consider the following portion of the diagram:
Given an element $\bar{c}$ of $H^n(\mathcal{C})$ , choose a representative $c\in \ker d_n\subset C^n$ . Since the vertical map from $B^n$ is surjective, choose $b\in B^n$ with $\beta_n(b)=c$ . Then $d_n(b)\in\ker\beta_{n+1}$ since the square commutes and $c\in\ker d_n$ ; thus $d_n(b)\in\im \alpha_{n+1}$ . Let $a\in A^{n+1}$ be the unique preimage of $d_n(b)$ . Note that $$ \alpha_{n+2}d_{n+1}(a)=
d_{n+1}\alpha_{n+1}(a) = d_{n+1}d_n(b)=0 $$ Since $\alpha_{n+2}$ is injective, $a\in\ker d_{n+1}$ . Then define $\delta_n(\bar{c})$ to be the equivalence class of $a$ in $H^{n+1}(\mathcal{A})$ . It is a straightforward diagram chase to verify that the result is independent of the choice of representative of $\bar{c}$ and of preimage of $c$ .
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"long exact sequence in cohomology" is owned by rm50.
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morphism of cochain complexes, map of cochain complexes, exact sequence of cochain complexes |
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Cross-references: independent, equivalence class, injective, preimage, surjective, element, homomorphism, proof, cohomology groups, exact sequence, short exact sequence, kernels, images, group homomorphisms, induces, diagram, maps, collection, morphism, cochain complexes
There are 2 references to this entry.
This is version 2 of long exact sequence in cohomology, born on 2009-10-09, modified 2009-10-09.
Object id is 11947, canonical name is LongExactSequenceInCohomology.
Accessed 310 times total.
Classification:
| AMS MSC: | 18G35 (Category theory; homological algebra :: Homological algebra :: Chain complexes) |
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Pending Errata and Addenda
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![$\displaystyle \xymatrix{ \cdots \ar[r] & A^n \ar[r]\ar[d]^{f_n} & A^{n+1} \ar[r... ...^{f_{n+1}} & \cdots \ \cdots \ar[r] & B^n \ar[r] & B^{n+1} \ar[r] & \cdots } $](http://images.planetmath.org:8080/cache/objects/11947/js/img1.png "$\displaystyle \xymatrix{ \cdots \ar[r] & A^n \ar[r]\ar[d]^{f_n} & A^{n+1} \ar[r... ...^{f_{n+1}} & \cdots \ \cdots \ar[r] & B^n \ar[r] & B^{n+1} \ar[r] & \cdots } $")
![$\displaystyle \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d]\ \cdots \ar[r] & A^... ...{n+1} \ar[r]^{d_{n+1}}\ar[d] & C^{n+2} \ar[r]\ar[d] & \cdots \ & 0 & 0 & 0 } $](http://images.planetmath.org:8080/cache/objects/11947/js/img3.png "$\displaystyle \xymatrix{ & 0 \ar[d] & 0 \ar[d] & 0 \ar[d]\ \cdots \ar[r] & A^... ...{n+1} \ar[r]^{d_{n+1}}\ar[d] & C^{n+2} \ar[r]\ar[d] & \cdots \ & 0 & 0 & 0 } $")