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PhD Thesis

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En la secuencia anterior sabemos que la suma y resta alternada de las dimensiones<br />

de los módulos nos debe dar cero. De aquí es sencillo calcular<br />

dim(H m−r+1 ((Lj(f1, . . . , fr+1))) = −dim(H m−r ((Lj(f1, . . . , fr)))<br />

H m−r (Lj(f1, . . . , fr+1)) − H m−r ((Lj−1(f1, . . . , fr+1) ≤m−r+2 )<br />

dim(H m−r+1 ((Lj(f1, . . . , fr)) ≤m−r+2 )) + dim(H m−r+1 (Lj−1(f1, . . . , fr+1))<br />

−dim(ker(i ∗ )).<br />

Un proceso de inducción finaliza la prueba.<br />

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