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    $\begingroup$ A perhaps more standard reasoning shows directly $\ell_\infty$ and $L_\infty(0,1)$ (both of which are $1$-injective) embed isometrically into each other. In fact, this is the usual way to show that these two spaces are isomorphic as Banach spaces, and it is not hard to see that they are not isometrically isomorphic. This approach avoids a discussion of injective envelopes. That $\ell_\infty$ embeds isometrically into $L_\infty(0,1)$ is obvious, and the other embedding follows from the fact that the predual $L_1$ of $L_\infty(0,1)$, being separable, is a quotient of $\ell_1$. $\endgroup$
    – Bill Johnson
    Commented Jul 5 at 18:04
  • $\begingroup$ Absolutely true; I was too carried away with the injective envelopes of separable Banach spaces that I totally missed $L_{\infty}$ within an arm's reach. $\endgroup$
    – Onur Oktay
    Commented Jul 5 at 19:22