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It is well known in narrow circles that the homological dimension (in the sense of relative Banach homology) of $c_0$-module $\ell_\infty$ is 2. As the corollary, this module is not projective. This proof is rather involved, its main ingridient is a lack of a right inverse for the mapping: $$ \Delta:c_0\;\hat{\otimes}\;c_0\to(c\;\hat{\otimes}\;c_0)\oplus(c_0\;\hat{\otimes}\;\ell_\infty): x\;\hat{\otimes}\;y\mapsto (x\;\hat{\otimes}\;y)\oplus(x\;\hat{\otimes}\;y) $$ in the category of left Banach $c_0$-modules.

I would like to see a more direct proof of non-projectivtity. The standard route would be to show that there is no right inverse $c_0$-morphism for the mapping $\pi:c \;\hat{\otimes}\; \ell_\infty\to \ell_\infty \colon a\; \hat{\otimes}\; x\mapsto a\cdot x$, where $c$ is the Banach space of convergent sequences.

Does anyone have an idea how to prove non-projectivity more or less directly?

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    $\begingroup$ Just to make the question self-contained... How does $c_0$ act on $c\widehat\otimes \ell^\infty$? $\endgroup$ – Matthew Daws Oct 15 '18 at 14:50
  • $\begingroup$ @MatthewDaws, simply the elementwise multiplication. $\endgroup$ – Norbert Oct 15 '18 at 16:04
  • $\begingroup$ Yes, but on which tensor factor? $\endgroup$ – Matthew Daws Oct 15 '18 at 16:11
  • $\begingroup$ @MatthewDaws, on the first one by the formula $a(b\;\otimes\; x)=ab\;\otimes\; x$, where $a\in c_0, b\in c, x\in\ell_\infty$. $\endgroup$ – Norbert Oct 15 '18 at 17:27
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I believe the following works...

Notice that $c=c_0 + \mathbb C1$ and so $$\newcommand{\proten}{\widehat\otimes} c\proten\ell^\infty = c_0\proten\ell^\infty + 1 \otimes \ell^\infty.$$ This is an isomorphism, maybe not isometric. Suppose, towards a contradiction, that there is a right inverse $T:\ell^\infty \rightarrow c\proten\ell^\infty$, so $T$ factors as $$ T(x) = T_1(x) + 1\otimes T_2(x)\qquad(x\in\ell^\infty), $$ where $T_1:\ell^\infty\rightarrow c_0\proten\ell^\infty$ and $T_2:\ell^\infty\rightarrow\ell^\infty$. That $T$ is a left $c_0$-module homomorphism means that $$ T(ax) = T_1(ax) + 1\otimes T_2(ax) = a\cdot T(x) =a\cdot T_1(x) + a\otimes T_2(x) \qquad (a\in c_0, x\in\ell^\infty). $$ Thus $T_2(a)=0$ for each $a\in c_0$ and $T_1(ax) = a\cdot T_1(x) + a\otimes T_2(x)$ for $a\in c_0, x\in\ell^\infty$. Finally, we should have that $\pi T(x)=x$, that is, $$ \pi_1 T_1(x) + T_2(x) = x \qquad (x\in\ell^\infty), $$ where $\pi_1:c_0\proten\ell^\infty\rightarrow\ell^\infty$ is the multiplication. Notice that $\pi_1$ takes value in $c_0$.

Then, for $a\in c_0$, as $T_2(a)=0$, we see that $\pi_1T_1(a)=a$. Thus $\pi_1T_1:\ell^\infty\rightarrow c_0\subseteq\ell^\infty$ is a projection, which is well-known not to exist. (This is Phillip's Lemma.)

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  • $\begingroup$ Thank you for your nice solution and my apologies for delayed reply. I looked through your solution and realized that it could be generalized to the following proposition. If a Banach $A$-module $X$ is projective, then $\operatorname{span}(A X)$ is complemented in $X$. $\endgroup$ – Norbert Oct 27 '18 at 1:07
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If $\ell^\infty$ was projective then any Banach limit $L:\ell^\infty \to \mathbb{C}$ would extend to a $c_0$-module morphism $\bar{L}:\ell^\infty \to c$ such that $L=\lim\circ \bar{L}$, but such does not exist.

Assume such $\bar{L}$ does exist. Consider $r=\bar{L}(1)-1\in c$ and observe that $\lim r=\lim\bar{L}(1)-1=L(1)-1=0$, thus $r\in c_0$. Let $I\subset \mathbb{N}$ be the set of indices $n$ for which $r_n=-1$ and let $V<\ell^\infty$ be the vector space of sequences supported on $I$. Note that $I$ is finite, thus $V$ is finite dimensional. For $s\in c_0$, $\bar{L}(s)=s\bar{L}(1)=sr+s$. It follows that $c_0\cap \ker\bar{L}=V$. But, as $V$ and $c$ are separable and $\ell^\infty$ is not, we can find $x\in \ker \bar{L}$ which is not in $V$. We consider the element $s=(1/n)\in c_0$ and observe that $sx$ is in $c_0\cap\ker\bar{L}$ but not in $V$. This is a contradiction.

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  • $\begingroup$ This a good proof from the first principles. Thank you. $\endgroup$ – Norbert Oct 26 '18 at 16:01

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