Injective Banach spaces, with morphisms as contractive linear maps, have been classically studied (and are $C(K)$ spaces with $K$ Stonian). But what about projectives?

So $P$ will be projective if given Banach spaces $E$ and $F$ and a quotient map (aka metric surjection) $\psi:E\rightarrow F$, given any contractive $\phi:P\rightarrow F$, we can lift this to a contractive $\varphi:P\rightarrow E$ with $\psi\varphi = \phi$. (If someone can make a nice commutative diagram, go ahead and edit this!)

**Claim:** The scalar field (say $\mathbb C$, but also works for $\mathbb R$) is not projective.

**Proof:** Let $f:c_0\rightarrow\mathbb C$ be the contractive functional $f(x) = \sum_{n=1}^\infty 2^{-n} x_n$ for $x=(x_n)\in c_0$. This induces an isometric isomorphism $c_0 / \ker(f) \cong \mathbb C$. So if $\mathbb C$ is projective, we can find a *contractive* $g:\mathbb C\rightarrow c_0$ with $fg=1$. That is, $g=(g_n)\in c_0$ is a norm-one vector with $f(g) = \sum 2^{-n} g_n =1$, which is impossible.

Question:Are there any projective Banach spaces?

It seems to me that the problem is insisting upon *contractive* morphisms. In the proof above, if we just need, for each $\epsilon>0$, to find $g$ with $\|g\|<1+\epsilon$, then this is no problem. Is anything known in this generalised setting? (It's easy to see that then $\ell_1(\Gamma)$ is always projective, if I allow myself this wiggle-room).