gives $a\cdot e\le 0$, hence $a\cdot e=0$, and hence $e=0$ by . This shows that $\Gamma_{\rho'}=\Gamma_{\rho,e}=\Gamma_\rho$ and $c_1(\rho')=c_1(\rho)+2e=c_1(\rho)$. The proof for $b^+=1$ is verbatim the same. Include $\kappa_\rho$, respectively $\kappa_{\rho'}$, as an argument of ${{\mathrm{SW}}}$ and use the fact that $\kappa_\rho=\kappa_{\rho'}$, because $[\rho]=[\rho']$.

The next corollary is a special case of a result by Conolly-Lé-Ono [@CLO]. It strengthens Corollary A in the simply connected case.

[**Corollary B.**]{} [*Let $M$ be a simply connected closed smooth four-manifold.*]{}

[**(i)**]{} [*Two symplectic forms on $M$ with the same first Chern class and inducing the same orientation on $M$ are homotopic as nondegenerate $2$-forms.*]{}

[**(ii)**]{} [*Two cohomologous symplectic forms on $M$ are homotopic as nondegenerate $2$-forms.*]{}

(Following Conolly–Lé–Ono [@CLO].) Assume $b^+=1$. (The case $b^+\ge2$ is easier.) Fix an orientation of $M$ and an integral lift $c\in H^2(M;{{\mathbb Z}})$ of the second Stiefel–Whitney class with $c^2=2\chi+3\sigma$. Let ${{\mathcal R}}^c$ be the set of nondegenerate $2$-forms on $M$ with first Chern class $c$ and compatible with the orientation. Then ${{\mathcal R}}^c$ has precisely two connected components, by . Let $\rho,\rho'\in{{\mathcal R}}^c$ be symplectic forms. Then $c_1(\rho')=c_1(\rho)=c$ and so $\Gamma_{\rho'}=\Gamma_\rho$, since $M$ is simply connected. By , $
{{\mathrm{SW}}}(M,{{\frak o}}_\rho,\kappa_\rho,\Gamma_\rho) = 1
= {{\mathrm{SW}}}(M,{{\frak o}}_{\rho'},\kappa_{\rho'},\Gamma_{\rho'}).
$ The 