The second fundamental form is intrinsic to the abstract Riemannian manifold

Define embeddings $P,C:U\to {\mathbb{R}}^3$ by $P(x,y)=(x,y,0)$ and $C(x,y)=(x,\cos y,\sin y)$. Their tangent pairs are $P_x=(1,0,0)$, $P_y=(0,1,0)$ and $C_x=(1,0,0)$, $C_y=(0,-\sin y,\cos y)$, so direct dot products give $P^{\ast }\bar{g}=C^{\ast }\bar{g}=d{x}^2+d{y}^2$. Thus $C\circ P^{-1}$ is an isometry from the planar strip $P(U)$ to the half-cylinder $C(U)$, realizing exactly the same abstract Riemannian manifold.

All second coordinate derivatives of $P$ vanish. Since [F2] identifies the ambient Euclidean covariant derivative with ordinary coordinate differentiation, [F1] gives ${\mathrm{I}\mathrm{I}}_P({\mathrm{\partial }}_x,{\mathrm{\partial }}_x)={\mathrm{I}\mathrm{I}}_P({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)={\mathrm{I}\mathrm{I}}_P({\mathrm{\partial }}_y,{\mathrm{\partial }}_y)=0$.

Along the cylinder let $N(x,y)=(0,\cos y,\sin y)$, a smooth unit normal. One has $C_{xx}=C_{xy}=0$ and $C_{yy}=(0,-\cos y,-\sin y)=-N$. Hence [F1]–[F2] give ${\mathrm{I}\mathrm{I}}_C({\mathrm{\partial }}_x,{\mathrm{\partial }}_x)={\mathrm{I}\mathrm{I}}_C({\mathrm{\partial }}_x,{\mathrm{\partial }}_y)=0$ but ${\mathrm{I}\mathrm{I}}_C({\mathrm{\partial }}_y,{\mathrm{\partial }}_y)=-N\mathrm{\ne }0$.

The isometry in step 1.1 identifies the two ordered orthonormal tangent frames, yet steps 2.1–2.2 identify a component that is zero for one embedding and nonzero for the other. Therefore $\mathrm{I}\mathrm{I}$ is not intrinsic.

The witness is nonempty, boundaryless, two-dimensional, and has positive-definite induced metric. Empty, zero-dimensional, and one-dimensional cases cannot invalidate this explicit two-dimensional counterexample to the universal claim. The open strip excludes the parameter endpoints $0,\pi$. Countable choice is inherited exactly through [A1] and [F1]; both embeddings and the cylinder normal are explicit, so no further selection occurs. No biconditional is asserted.