The cut locus of a point is always a smooth hypersurface

Take the unit round sphere $M=S^2\subset {\mathbb{R}}^3$ with north pole $p=(0,0,1)$. [F1, F2, F3] By [F3] it is a nonempty boundaryless Riemannian manifold, and [F2] makes it connected. The all-real great-circle formulas in [F3] show that it is geodesically complete; hence it is metrically complete by [F1]. Thus this is within the domain of the cut-locus definition under its declared ${\mathrm{A}\mathrm{C}}_{\omega }$ assumption.

For $q\in S^2$, define $\theta (q):=\arccos ⟨p,q⟩\in [0,\pi ]$. [step 1.1, F3, F4, F5, F6, F8, F11] There is a smooth path from $p$ to $q$ of length $\theta (q)$. If $q=p$, use the constant path, which has length $0=\theta (p)$. If $q=-p$, let $u=(1,0,0)$ and use $\beta (s)=\cos (s)p+\sin (s)u$ on $[0,\pi ]$; then $\beta (0)=p$ and $\beta (\pi )=-p$. Otherwise $0<\theta (q)<\pi$. Since cosine is strictly decreasing on $[0,\pi ]$ [F5], we have $\mathrm{\mid }\cos \theta (q)\mathrm{\mid }<1$, so [F11] gives $\sin \theta (q)\mathrm{\ne }0$. Set $u:=\frac{q-\cos (\theta (q))p}{\sin (\theta (q))}$ Then $⟨p,q-\cos (\theta (q))p⟩=0,\mathrm{\mid }q-\cos (\theta (q))p{\mathrm{\mid }}^2=1-{\cos }^2(\theta (q))={\sin }^2(\theta (q)),$ so $u$ is a unit vector orthogonal to $p$; use $\beta (s)=\cos (s)p+\sin (s)u$ on $[0,\theta (q)]$, which ends at $q$ by the definition of $u$. In each nonconstant case [F3] gives this great-circle geodesic. Orthonormality of $p,u$ and [F11] give $\mathrm{\mid }\beta (s){\mathrm{\mid }}^2={\cos }^2(s)+{\sin }^2(s)=1$, while $\dot{\beta }(0)=u$; thus $u$ is tangent to $S^2$ at $p$. Its derivative is $\dot{\beta }(s)=-\sin (s)p+\cos (s)u$, and the same calculation gives $\mathrm{\mid }\dot{\beta }(s){\mathrm{\mid }}_g^2={\sin }^2(s)+{\cos }^2(s)=1$. Its length is therefore ${\int }_0^{\theta (q)}1ds=\theta (q)$ by [F4], [F8], [F11]. Taking the infimum in [F4] gives $d_g(p,q)\le \theta (q)\mathrm{.}$

For the reverse inequality, let $\alpha :[a,b]\to S^2$ be any piecewise $C^1$ path from $p$ to $q$. [step 1.1, F4, F5, F6, F7, F8] If $a=b$, then $p=q$ and the desired lower bound is $0\le L_g(\alpha )$. Otherwise fix $0<\epsilon <1$ and, on each smooth piece, set $z(t):=⟨p,\alpha (t)⟩,h_{\epsilon }(t):=\arccos ((1-\epsilon )z(t))\mathrm{.}$ Since $\mathrm{\mid }z(t)\mathrm{\mid }\le 1$, the argument of $\arccos$ stays strictly between $-1$ and $1$. Also $⟨\alpha ,\dot{\alpha }⟩=0$, so $z^{\mathrm{\prime }}=⟨p-z\alpha ,\dot{\alpha }⟩,\mathrm{\mid }p-z\alpha {\mathrm{\mid }}^2=1-z^2\mathrm{.}$ By [F6], $\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }\le \sqrt{1-z^2}\mathrm{\mid }\dot{\alpha }\mathrm{\mid }$. The chain rule [F7] and derivative formula [F5] now give $\mathrm{\mid }{h}_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }=\frac{(1-\epsilon )\mathrm{\mid }{z}^{\mathrm{\prime }}\mathrm{\mid }}{\sqrt{1-(1-\epsilon {)}^2{z}^2}}\le \mathrm{\mid }\dot{\alpha }\mathrm{\mid },$ because $1-(1-\epsilon {)}^2{z}^2\ge (1-\epsilon {)}^2(1-z^2)$. The functions $h_{\epsilon }^{\mathrm{\prime }}$ and $\mathrm{\mid }\dot{\alpha }\mathrm{\mid }$ are continuous on each closed smooth piece, including their one-sided endpoint values, so they are integrable by [F8]. Integrating $-\mathrm{\mid }\dot{\alpha }\mathrm{\mid }\le h_{\epsilon }^{\mathrm{\prime }}\le \mathrm{\mid }\dot{\alpha }\mathrm{\mid }$, and applying the fundamental theorem and integral monotonicity and linearity [F8], bounds the absolute change of $h_{\epsilon }$ on that piece by its length. Summing over the finitely many pieces and using the triangle inequality yields $\mid \arccos ((1-\epsilon )⟨p,q⟩)-\arccos (1-\epsilon )\mid \le L_g(\alpha )\mathrm{.}$ As $\epsilon \downarrow 0$, continuity of $\arccos$ [F5] gives $\theta (q)\le L_g(\alpha )$.

By Steps 2.1 and 2.2, every piecewise $C^1$ path from $p$ to $q$ has length at least $\theta (q)$, and one such path has length exactly $\theta (q)$. [step 2.1, step 2.2, F4] Taking the infimum in [F4] proves the exact formula $d_g(p,q)=\arccos ⟨p,q⟩\mathrm{.}$

Fix a unit $u\in T_p{S}^2$. The round-sphere and radial-geodesic formulas give ${\gamma }_u(t)=\cos (t)p+\sin (t)u={\exp }_p(tu)$ [step 3.1, F3, F10] Its inner product with $p$ is $\cos t$. Thus [F5] and step 3.1 give $d_g(p,{\gamma }_u(t))=t$ for $0\le t\le \pi$. For $t>\pi$, $d_g(p,{\gamma }_u(t))=\arccos (\cos t)\le \pi <t$. Consequently the minimizing positive radial times are exactly $(0,\pi ]$, so by [F10] the cut time is $c_p(u)=\pi$ for every unit $u$. By the finite-endpoint clause in Cut point and cut locus of a point, $\mathrm{Cut}(p)=\{{\gamma }_u(\pi ):u\in S_{p}M\}=\{-p\}\mathrm{.}$ The set is nonempty since $(1,0,0)\in S_{p}M$.

In a chart centered at $-p$, the singleton $\{-p\}$ is an embedded zero-dimensional submanifold [step 4.1, F9]. Indeed, take a chart around $-p$ and translate its coordinates so that $-p$ maps to the origin. In any open chart image containing that origin, intersection with ${\mathbb{R}}^0\times \{0\}=\{0\}$ is just the singleton. But if $\{-p\}$ were an embedded $1$-dimensional submanifold, a slice chart at $-p$ would identify its one-point intersection with the intersection of an open subset of ${\mathbb{R}}^2$ and a coordinate line; that latter intersection contains an interval and cannot be a singleton. By [F9], a hypersurface in $S^2$ must have dimension $2-1=1$. Thus $\mathrm{Cut}(p)=\{-p\}$ is not a smooth hypersurface, refuting the universal claim.

Empty and zero-dimensional cases: the empty manifold has no base point; in dimension zero there are no unit tangent vectors [step 3.1, step 4.1, step 5.1, F1, F9, F10]. so the cut-point definition gives no cut points; the witness therefore requires positive dimension. Dimension one is not refuted by this example, since a singleton there has codimension one. The distance formula includes $q=p$ with $\theta =0$; on the witness every radial segment minimizes at $t=0$ and through the endpoint $t=\pi$, while all $t>\pi$ fail strictly. The antipodal direction uses the fixed vector $(1,0,0)$, and the general direction is explicitly determined by $q$, so there is no choice from a family. The only axiom used is the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ in [F1] and the cut-time interfaces; no full Axiom of Choice is invoked. This is a one-way counterexample, not an iff claim.