Zeros of the exact multiple-shooting residual are exactly BVP solutions

If the residual vanishes, each row gives ${\varphi }_i(t_{i+1};s_i)=s_{i+1}$. Define $y$ on each interval by the corresponding segment solution. At every interior mesh point, the left endpoint value equals the next initial state, so the pieces are continuous; the last residual row supplies $B(y(a),y(b))=0$.

At a mesh point $t_i$, both one-sided derivatives equal $f(t_i,y(t_i))$, since each $C^1$ segment solves the same ODE and the joined values agree. Thus the pieces join to a $C^1$ solution on $[a,b]$; the endpoint pair lies in $D_B$ and the last residual row gives its boundary equation, so it is a BVP solution.

Conversely, let $y$ be a BVP solution with $y(t_i)=s_i$ for every $0\le i\le N$, keeping the given tuple fixed. On each mesh interval its restriction and the given segment IVP solution have initial state $s_i$, so the assumed uniqueness identifies them. Thus ${\varphi }_i(t_{i+1};s_i)=y(t_{i+1})=s_{i+1}$ for $0\le i<N$, and $B(s_0,s_N)=B(y(a),y(b))=0$. All residual rows vanish.