Split exact sequences of vector spaces

Asked Jun 28 '15 at 14:10

Active Jun 28 '15 at 15:36

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The question is on page 2, exercise 1.1.3.

For the proof that $\{ C_n \}$ is a chain complex I only need to show that $(i\circ p)\circ (i\circ p) = 0$ where $i$ is the inclusion map, and $p$ is the projection, if I am not mistaken this follows from: $ (i\circ p)\circ (i\circ p) = i\circ p \circ i =0 $ since $p^2 =p $ and $i$ commutes with $p$ or so I think. If I am mistaken here, let me know how to change it?

I don't know though how to prove that every chain complex of vector spaces is isomorphic to a complex of this form.

The question from the book:

Choose vector spaces $\{ H_n , B_n \}_{n\in \mathbb{Z}}$ over a field, and set $C_n = B_n \oplus H_n \oplus B_{n-1} $. Show that the projection-inclusions $C_n \rightarrow B_{n-1} \subset C_{n-1}$ make $\{ C_n \}$ into a chain complex, and that every chain complex of vector spaces of vector spaces is isomorphic to a complex of this form.

edited Jun 28 '15 at 15:36

grg

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asked Jun 28 '15 at 14:10

MathematicalPhysicist

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You should really copy the exercise statement here, preferably in a quotation block. Also, I'm not sure about the legality of linking to a copy of Weibel's book. – A.P. Jun 28 '15 at 14:33
Iv'e added the question in quotation marks, hope you can help me with the question. – MathematicalPhysicist Jun 28 '15 at 14:44
2

"$p^2$" does not make sense since the range of $p$ is not even equal to the domain of $p$. The answer is simply that $p \circ i = 0$. For the second part of the question, notice that you are working over a field (thus there are bases). – Najib Idrissi Jun 28 '15 at 14:54
@NajibIdrissi Why does $p\circ i =0$? – MathematicalPhysicist Jun 29 '15 at 05:39
By definition... Just write it down! – Najib Idrissi Jun 29 '15 at 08:06
So wait a minute is it right that $p$ commutes with $i$? As for to show this, I believe that we first have the inclusion $i$ goes from $B_{n-1}\subset C_{n-1}$ and then the projection of $C_{n-1}$ into $B_{n-2}$, why does it vanish? – MathematicalPhysicist Jun 29 '15 at 08:25
If $p(x,y,z) = z$ and $i(x) = (x,0,0)$, why is $p \circ i = 0$...? I cannot do everything for you... – Najib Idrissi Jun 29 '15 at 11:11
Ah, ok. Thanks, I understnad it now. – MathematicalPhysicist Jun 29 '15 at 11:13

Split exact sequences of vector spaces

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