Does the Implicit Function Theorem in Banach spaces holds if the differential is only one-to-one (not onto!)?

Question

Is the Implicit Function Theorem in the following form correct:

Let $V_1,V_2,W$ be Banach spaces, and $Ω⊂V_1×V_2$ an open subset containing $(x_0,y_0)$. Let consider a continuously differentiable map $f:Ω→W$ with $f(x_0,y_0)=0$ and s.t. the derivative on the second component

$D_2f(x_0,y_0):V_2\ni y↦Df(x_0,y_0)(0,y)∈W$ is one-to-one (but not necessarly onto).

Then there exists an open set $Ω_1×Ω_2⊂Ω$ around $(x_0,y_0)$ and a unique map $g:Ω_1→Ω_2$ s.t $f(x,g(x))=0$ for all $x∈Ω_1$.

Indeed, the standard strategy would be to consider

$L \equiv D_2f_{(x_0,y_0)}$ and $ \Phi(x,y)\equiv y - (L^{-1}\circ f)(x,y). $

However, the map $\Phi$ is not defined everywhere. Indeed, it is defined only on

$$f^{-1}\bigg(L\big(V_2\big)\cap f(\Omega)\bigg).$$

I look forward to reading your comment, possibly references where I can find a proof if that statement is correct. Thanks in advance

Answer 1

No, it is not correct. Simple counterexamples can be obtained as follows. Let $V_1=W$ and $V_2$ be any Banach spaces such that the identity is continuous but not surjective $V_2\to W$. Then $f$ defined by $(x,y)\mapsto x+y$ is even smooth with $\partial_2 f(x_0,y_0)$ injective $V_2\to W$ and there cannot be any $g,\Omega_1,\Omega_2$ as required since we would have $g(x)=-x$ for all $x\in\Omega_1$ and further $W\subseteq V_2$.