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For my quetion in MO is $\forall X$, $X^{**}$=X$\oplus Y$ for a $Y$ another set I am not really sure in Thomas answer why the first assumption saying that such a $Y$ exist iff the sequence $0 \to X \to \varphi(X^∗)^∗ \to \eta \mathrm{coker}\varphi \to 0$ splits?

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No. This question has been asked and answered on MathOverflow. – Alex Becker Sep 16 '12 at 4:45
3 fact, it looks like you asked this question there, and I answered it. – Alex Becker Sep 16 '12 at 4:50
What is true is that every dual space is complemented in its bidual. All the spaces you listed are dual spaces. The space $c_0$ is not complemented in its bidual Phillips's lemma: $(c_0)^{\ast\ast} = \ell^{\infty} \ncong c_0 \oplus Y$ for any $Y$. – t.b. Sep 16 '12 at 4:52
If there are some points that are unclear in Alex's answer from MO, perhaps inquire about these specific points instead of asking the same question. – arjafi Sep 16 '12 at 5:07
I personally would be interested to know a further characterization of when the sequence does and does not split, based on properties of $X$. To add, it seems that the splitting condition is just a restatement of the original issue but in different terms. – Nick Alger Sep 16 '12 at 6:56

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