Consider the embeddings $
\psi_i:=g\circ\widehat{\psi}_i:B^4_{r_i+\delta}\to{{{{\mathbb C}}\mathrm{P}}}^2.
$ Composing $g$ with a suitable Hamiltonian isotopy we may assume without loss of generality that $\psi_i(0)=p_i$ and that $\psi_i$ agrees with ${{\varphi}}_i$ near the origin. Thus $g(\widehat{N}_i) = \psi_i(B^4_{r_i+\delta})$ is a neighborhood of $p_i$ and the complement of their union is symplectomorphic to a subset of $M$ under $\pi^{-1}$. The formula in the proof of [@MS1 Lemma 7.15] gives rise to a diffeomorphism $f:M\to M$ such that $f(C_i) = S_i = \pi^{-1}(p_i)$, $f(N_i) = \pi^{-1}(g(\widehat{N}_i))$ and $f|_{M\setminus(N_1\cup N_2)}=\pi^{-1}\circ g\circ\widehat{\pi}$. The diffeomorphism $f$ satisfies $$f_*L^c = L,\qquad f_*E^c_1 = E_1,\qquad f_*E^c_2 = E_2.$$ The symplectic form $\rho:=(f^{-1})^*{{\omega}}$ agrees with $\rho_0$ on $M\setminus\pi^{-1}(g(\widehat{N}_1)\cup g(\widehat{N}_2))$. It can be joined to $\rho_0$ by a path of symplectic forms via a scaling argument. If $c_1({{\omega}})=c_{{\mathrm{std}}}$ one can choose $E^c_i=E_i$ and then $f$ induces the identity on homology. This proves (V) and (VI).

[**7.**]{} The proof of (VII) hinges on the fact that  holds for the two point blowup of ${{{{\mathbb C}}\mathrm{P}}}^2$ and symplectic forms with the standard first Chern class (i.e. two cohomologous symplectic forms with first Chern class $c_{{\mathrm{std}}}$ are homotopic if and only if they are isotopic). This was proved by McDuff in [@MCDUFF4; @MCDUFF5]. Thus 