linear map $J(X,\tilde \Sigma,\nu)$ is equal to the old one if $g$ belongs to $D^{+}$ or to $M$. If $g$ belongs to $D^{-}$, then $J(X,\tilde \Sigma,\nu)$ is zero.

This is an obstructed gluing problem of just the same sort as in the previous proposition. Once again, the branched doubee cover of $S^{4}$ over the surface $R=D^{+}\cup M$ has $b^{+}=1$, and the calculation proceeds as before. (One can alternatively derive this proposition from the previous one by showing that $\Psi$ is a connect sum $\Psi^{-}_{2} \#_{t,t}\Psi_{2}$ at the tetrahedral points.)

Topology of the composite cobordisms
====================================

As stated in the introduction, the proof of the exact triangle (Theorem \[thm:exact-triangle\]) is very little different from the proof of a corresponding result for the $\SU(2)$ instanton knot homology $\Isharp$. The first step is to understand the topology of the composites of the cobordisms $\Sigma(L_{i+1}, L_{i})$. We shall abbreviate the notation for these cobordisms to just $\Sigma_{i+1,i}$.

Figure \[fig:mobiusshadedwithfeetfoamtwodisks\] shows (schematically) the composite of three consecutive foam cobordisms, from $L_{3}$ to $L_{0}$. The indices are interpreted cyclically, so $L_{3}$ and $L_{0}$ are the same web in $Y$. To explain the picture, in the foam $\Sigma_{1,0}$ pictured in Figure \[fig:cobordism\], there are two half disks 