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



Does the Implicit Function Theorem in Banach spaces holds if the differential is only one-to-one (not onto!)?
**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 $W\subseteq V\_2$.


