I'm looking for a book containing the proof that for every Banach space E there is an index I so that E is a quotient space of $\ell_1(I)$. If I can't find the book on google books, it would be great if you could give me the page number, because my paper is due tomorrow and I just need it for a reference. Thank you!
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Is this the reference you needed? – Tomasz Kania Oct 18 '14 at 19:19
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Lemma 1.4.a in:
J. M. F. Castillo and M. Gonzalez, Three-space problems in Banach space theory, Springer Lecture Notes in Math. 1667, 1997.
Proof. Let $E$ be a Banach space and let $\{x_i\colon i\in I\}$ be a dense subset of the unit ball of $E$. Define a map $Q\colon \ell_1(I)\to E$ by
$$Q ( (a_i)_{i\in I} ) = \sum_{i\in I} a_i x_i\quad (a_i)_{i\in I} \in \ell_1(I).$$
Then $Q$ is the desired surjection. $\square$
Tomasz Kania
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Just one comment: in their book, they pick a dense subset of the unit ball but they need a dense subset of the unit sphere for the map to be an isometry. – Filippo Alberto Edoardo Apr 11 '22 at 07:37
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Which map are you speaking about, precisely? What I meant is simply that in Lemma 1.4.a op. cit they claim that $X$ is linearly isometric to some quotient of $\ell^1(I)$ for some set $I$. The map $Q$ they construct sends $e_\alpha\in\ell^1(I)$ to $x_\alpha$ and this will not be an isometry unless $\Vert x_\alpha\Vert =1$. – Filippo Alberto Edoardo Apr 11 '22 at 12:11
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@FilippoAlbertoEdoardo, this map $Q$ has a kernel so can't be isometric. – Tomasz Kania Apr 11 '22 at 19:07
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You need the Banach space to be separable, I think.
You can find it as Theorem 4.6 in these notes by Piotr Hajlasz.
Batman
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Of course, $E$ need not be separable in this case $I$ is just uncountable. – Tomasz Kania Sep 29 '14 at 13:35