Euler number of a clutched bundle as the clutching degree

Put $s_{+}=v_0$ on the upper hemisphere. By the upper-to-lower transition, the lower boundary value must be $s_{-}(a)=g(a)v_0$. Extend this value to a smooth map $F:D_{-}^4\to {\mathbb{R}}^4$, constant in the radial coordinate near the boundary and zero near the centre, by a smooth radial cutoff. It agrees with the constant upper section through the equatorial product charts. The base orientation on $D_{-}^4$ is its standard coordinate orientation, and its coordinate boundary orientation is the source orientation fixed in the statement.

The map $F$ is nonzero on a boundary strip. Choose a smooth cutoff $\rho$ equal to one on the compact complement of that strip, supported away from the boundary, and with its transition region inside the strip. By [L6] choose a sufficiently small regular value $t$ of $F$ on the open disk. Then $F_t=F-\rho t$ has no zeros in the transition strip, while every remaining zero lies where $\rho =1$ and has invertible derivative $DF$. Thus the section with $s_{+}=v_0$, $s_{-}=F_t$ is smooth and transverse to zero, with a finite zero set in the lower open disk.

By [L2] its signed zero count is $⟨e(E),[S^4]⟩$. On the positively oriented lower disk each local contribution is $\mathrm{sign}\det D{F}_t$, the index of the coordinate vector field $Y=F_t$ by [L4]. Hence [L3] gives $⟨e(E),[S^4]⟩=\mathrm{deg}(a\mapsto Y(a)\mathrm{/}\mathrm{\mid }Y(a)\mathrm{\mid })=\mathrm{deg}(a\mapsto g(a)v_0)$, since the boundary was fixed and $v_0$ is unit. Scaling a nonzero $v_0$ to unit length gives the displayed general formula.

For either basic clutching choose $v_0=1$. Both boundary maps are then $a\mapsto a$, of degree $+1$ by [L5]. For any other unit $v_0$, a path from $1$ to $v_0$ in $S^3$ gives a homotopy of the boundary maps, so their degree is unchanged. This proves both calibrations in the stated orientation convention.