This is a follow-up to my latest question: If $f : \Omega \subset \mathbb{C} \to X$ is an injective Banach-valued holomorphic function, can $f'$ ever have a zero?. It was answered concisely and swiftly, but I would like to turn part of what I had written into its own question. This question might thus feel similar, but the focus will be different this time.
Let $X$ be a complex Banach space, and consider $f : \Omega \subset \mathbb{C} \to X$ a (hence Banach-valued) holomorphic function on an open set $\Omega$. I'll denote by $X^*$ the (topological) dual of $X$.
I would like to (dis)prove the following claim:
If $f$ is injective and $f'$ has no zeros in $\Omega$, does there necessarily exist a $\Lambda \in X^*$ such that $\Lambda \circ f$ is injective?
When $X = \mathbb{C}$ and we go back to the standard theory of complex-valued holomorphic functions, the $f'$ part of the assumption is redundant, as it is a consequence of the open mapping theorem that if $f$ is injective, then $f'$ has no zeros on $\Omega$, and even that $f$ has a holomorphic inverse $f^{-1} : f(\mathbb{C}) \to \Omega$.
However, thanks to Moishe Kohan's simple example $z \in \mathbb{C} \mapsto (z^2, z^3) \in \mathbb{C}^2$, which is injective but whose derivative has a zero at $0$, it is not the case in a generic $X$ that injectivity implies having no critical points.
To conclude, the claim is trivially true for $X = \mathbb{C}$ since the $\Lambda$s are of the form $z \mapsto az$.
Now, my claim did not fall from the sky: it is motivated by the part of my earlier question where I discussed my efforts, which I'll mostly copy-and-paste to keep this question self-contained:
I had the idea of attempting to look at the open discs $D$ such that $\bar{D} \subset \Omega$, and trying to make a finite-intersection-property type proof on each $\bar{D}$ compact to show that if the $\Lambda \circ f$, $\Lambda \in X^\ast$ are all non-injective on $\bar{D}$, then $f$ is non-injective on $\bar{D}$ [corrected from my last question, where I put $D$] or $f'$ has a zero on $\bar{D}$. Finally, this means in particular that $f$ is non-injective on $\Omega$ or that $f'$ has a zero in $\Omega$, and we would be done.
First, define the following function: $$g : (z,w) \in \Omega^2 \mapsto \begin{cases}\displaystyle\frac{f(z) - f(w)}{z - w} \,&\text{if } z \neq w \\ f'(z) \,&\text{if } z = w \end{cases} \in X$$
$g$ can be shown to be continuous just like in the complex-valued case.
Now let $\mathcal{P}_{fin}(X^\ast)$ be the set of all finite subsets of $X^\ast$, and pick $L \in \mathcal{P}_{fin}(X^\ast)$. Define the following set:
$$E_L := \left\{(z,w) \in \bar{D}^2 \mid \forall \Lambda \in L,\, \Lambda(g(z,w)) = 0\right\}$$
Each $E_L$ is closed by continuity of the $\Lambda$s and of $g$, and contained in $\bar{D}$ compact. Moreover, we have, for $(L_1, \dots, L_d) \in \mathcal{P}_{fin}(X^\ast)^d$:
$$\begin{cases}\bigcap_{k = 1}^d E_{L_k} = E_{\bigcup_{k = 1}^d L_k} \\ \bigcup_{k = 1}^d L_k \in \mathcal{P}_{fin}(X^\ast)\end{cases}$$
Hence, it would only remain to show that all $E_L$s are non-empty to have that $(E_L)_{L \in \mathcal{P}_{fin}(X^\ast)}$ has the finite intersection property in $\bar{D}$ compact, and thus the intersection of all the $E_L$s would be non-empty. But that intersection is exactly the set of the zeros of $g$ on $\bar{D}^2$, therefore $g$ would have a zero on $\bar{D}^2$, which would imply that $f$ is non-injective or that $f'$ has a zero.
The main issue here is that I do not yet see a way to show that the $E_L$s are non-empty... the assumption on the $\Lambda$s means that the $E_{\{\Lambda\}}$s are non-empty, so I'm guessing that induction on the size of the $L$s would be the way to go?
I'm probably missing a fairly simple counterexample yet again, but that's how life goes.
Any help is appreciated, and feel free to edit or re-tag if/when necessary.
It would seem I once again overlooked a simple counterexample (thank you PhoemueX!):
Consider $X = \mathbb{C}^2$, and the function $f : z \in \Omega := \mathbb{C} \setminus \{0\} \mapsto (z^2, z^3)$, i.e. Moishe's example from my last question but without the origin. It is holomorphic, injective and $f'$ has no zeros on $\Omega$.
Now, all the linear functionals on this space are of the form $\Lambda_{a_1,a_2}(z_1, z_2) \mapsto a_1 z_1 + a_2 z_2$ for some $(a_1, a_2) \in \mathbb{C}^2$, hence $\Lambda_{a_1, a_2}(f(z)) = a_1 z^2 + a_2 z^3$. These are polynomials of degree $-\infty$, $2$ or $3$, thus never injective, and my claim was false.