For closed 4 manifold X, we consider the derivative of the Seiberg-Witten functional, i.e.
$$\Omega^1\_2(X;\sqrt{-1}\mathbb R)\oplus\Gamma\_2(S^+)\overset{D}{\to}\Omega^2\_{+,1}(X;\sqrt{-1}\mathbb R)\Gamma\_1(S^-),$$
where 


1. $\Omega^i\_j(X;\sqrt{-1}\mathbb R)$ means $\sqrt{-1}\mathbb R$-valued $i$-forms space with Sobolev $L^{2,j}$ norm;
2. $\Omega^2\_\pm$ means space of the self/anti-self dual parts of 2-forms ;
3. $\Gamma\_i(S^\pm)$ means the sections of $S^\pm$-spinor space with Sobolev $L^{2,i}$ norm;
4. $$D\_{A,\psi}=\left(\begin{array}{cc}d^+&-Dq\_\psi\\
\cdot\frac12\psi&D\_A\\
\end{array}\right)$$
here 


	* $Dq\_\psi(\eta)=\psi\otimes\eta^\*+\eta\otimes\psi^\*-\frac{<\eta,\psi>+\overline{<\eta,\psi>}}2Id$,
	* $D\_A$ means the the Dirac operator w.r.t. the connection $A$,
	* $\cdot$ means the Clifford action,
	* $d^+$ means the differential $d$ and orthogonal project to the self-dual part.


Q: Does $D$ admits a right inverse $R$, $DR=Id$?



Right inverse of the Seiberg-Witten functional
This is found in any reference on SW-theory. I am implicitly assuming we have perturbed the SW-equations, with generic perturbation. The linearization (which includes the gauge-action) at a given SW-solution forms an elliptic complex. This operator is surjective, hence admits a right inverse.


I should clarify: the point of the perturbation is for all SW-solutions to be *regular*, which by definition means that the linearized operator is surjective at all SW-solutions.


