Is the following result true: the Hilbert space $\ell^{2}\left(2^{\Gamma}\right)$ is a quotient of $\ell^{\infty}\left(\Gamma\right)$ for any
uncountable $\Gamma$ ? [I think it is, but cannot remember where I saw it, long time ago.] I would be very grateful for any (freely available, if possible) reference (Pelczynski ? Rosenthal ?).



Nonseparable Hilbert spaces as quotients of spaces of bounded functions
I don't know who first observed this (maybe Archimedes?) but it is true because $C(\{0,1 \}^\Gamma)$ is a quotient of $\ell\_1^\Gamma$ and hence $\ell\_1(2^\Gamma)$ embeds into $\ell\_\infty(\Gamma)$. 


@Ady 


Here is a more serious answer to your question. Take a quotient map $Q$ from $\ell\_1(2^\Gamma)$ onto $C([0,1]^{2^\Gamma})$ and extend to a norm one mapping $T$ from $\ell\_\infty(\Gamma)$ into some injective space $Z$ that contains $C([0,1]^{2^\Gamma})$ (you cannot extend $Q$ to an operator from $\ell\_\infty(\Gamma)$ into $C([0,1]^{2^\Gamma})$ because, e.g., $C([0,1]$ is not a quotient of $\ell\_\infty$). Use partitions of unity to get a net $(P\_a)$ of norm one finite rank projections on $Z$ taking values in $C([0,1]^{2^\Gamma})$ and whose restrictions to $C([0,1]^{2^\Gamma})$ converge strongly to the identity. A weak$^\*$ cluster point of $(P\_a^\* T^\*)$ gives an isometric embedding of the dual of $C([0,1]^{2^\Gamma})$ (which contains $L\_1([0,1]^{2^\Gamma})$) into the dual of $\ell\_\infty(\Gamma)$. Thus if $Y^\*$ is any reflexive subspace of $L\_1([0,1]^{2^\Gamma})$, such as $\ell\_2(2^\Gamma)$, then $Y$ is isometric to a quotient of $\ell\_\infty(\Gamma)$.


