Injective holomorphic/smooth map into a topological vector space vs immersion. In the finitely dimensional case it follows from the constant rank theorem that a smooth map between smooth manifolds, which is not an immersion on an open set, cannot be injective. I need a "semi-infinite" dimensional analogue of this fact.
Namely, let $U$ be a domain in $\mathbf{C}^{d}$, let $E$ be a topological vector space and let $\varphi:U\to E$ be holomorphic, such that the rank of $D\varphi_{x}$ is less than $d$ for every $x\in U$. Does it follow that $\varphi$ is not injective?
I understand that the constant rank theorem does exist for infinite dimensions, but it has a lot of technical conditions, which I suspect to be redundant if the domain is finitely dimensional. In particular, I am curious if we need $E$ to be good, e.g. isomorphic to a Banach space.
In fact, I can prove my claim; however I hope for a reference to a constant rank theorem or something else, that would just make my life simple =)
 A: The constant rank theorem is generally proved by gluing the submersion theorem and the immersion theorem : locally a constant rank map is obtained as $f= i\circ s$, where $i$ is an immersion and $c$ a submersion. 
This work in your case. Let $r<d$ be the rank of $\phi$ on some open set $\Omega$ where it is constant and maximal. 
For every continuous linear form on $E$ say $l\in E^*$, $\varphi ^*dl = d\varphi ^*l$ is an (exact) differential form on $M$, and at each point $x$ the subspace   $K _x$ of the dual of $\bf R^d$ (in fact of $T_x^*\Omega$) generated by these forms is of constant dimension $r$. We can therefore find a finite subset $l_1,..., l_{r}$ such that these forms are linearly independent in a neighborhood $\Omega '$ of a given point.
Now we can argue like in finite dimension.
The map $s$ defined by $s : \Omega ' \to \bf R^{r}$ defined by $s(x)= (l_1\circ \phi,...l_{r}\circ \phi )$ is a submersion (in particular is not injective).
In order to prove the "constant rank theorem" in this context, it is enough to check that $\varphi$ is constant along the fibers of $s$. To this end, note that
the kernel of the derivative $D\varphi $ at a point $x$,   $K_x$, is the the finite intersection of the kernels of the forms $\phi ^*dl _i$ at each point on $\Omega '$, this is the same as the  kernel of $Ds$
