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quasar987

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I do not see how the conclusion follows from the fact that we can "choose" {(1,0), (1,-1)} as a basis for

**Z**²...

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- #1

quasar987

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I do not see how the conclusion follows from the fact that we can "choose" {(1,0), (1,-1)} as a basis for

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matt grime

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How do you compute the quotient

[tex]\frac{\ker(d_n)}{\text{im}(d_{n+1})}[/tex]

? If you can express [tex]\ker(d_n)[/tex] and [tex]\text{im}(d_{n+1})[/tex] in terms of the same basis, then modding out is straight forward. That's why Wiki is choosing a non-standard basis for**Z**². Why don't you write out the sequence and the maps?

[tex]\frac{\ker(d_n)}{\text{im}(d_{n+1})}[/tex]

? If you can express [tex]\ker(d_n)[/tex] and [tex]\text{im}(d_{n+1})[/tex] in terms of the same basis, then modding out is straight forward. That's why Wiki is choosing a non-standard basis for

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- #4

quasar987

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Actually, there is no need to talk about basis here since Im(alpha) is clearly just 2

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