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?



Explicit proof that $c_0$-module $\ell_\infty$ is not projective
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](https://math.stackexchange.com/questions/132520/complement-of-c-0-in-ell-infty).)


