Geodesics, the Exponential Map, Completeness, and Hopf–Rinow

The interval gives the obstruction. Its Euclidean Christoffel symbol is zero in the interior, so an affinely parametrized geodesic with initial data $\gamma (0)=1\mathrm{/}2$ and ${\gamma }^{\mathrm{\prime }}(0)=1$ must locally be $\gamma (t)=1\mathrm{/}2+t$. It remains inside $I$ only for $-1\mathrm{/}2\le t\le 1\mathrm{/}2$ and cannot be continued as that solution for all real times while taking values in $I$.

Nevertheless [F2] makes $I$ a complete metric space. Thus the implication “metric completeness implies two-sided geodesic completeness” would be false if arbitrary manifold boundaries were silently admitted. The boundaryless convention in [F1] prevents this mismatch. The two endpoints and outward direction are explicit; the zero-dimensional case has only constant geodesics, the empty case is vacuous, and no selection or choice principle is used.

The velocity ${\gamma }^{\mathrm{\prime }}$ is a section of ${\gamma }^{\ast }TM$, so [F2] makes $D_t{\gamma }^{\mathrm{\prime }}$ well defined and intrinsic. The equation therefore compares vectors in $T_{\gamma (t)}M$ and is independent of any chart or extension of the velocity field.

If $\gamma (t)=p$ is constant, then ${\gamma }^{\mathrm{\prime }}$ is the zero section and [F2] gives $D_t{\gamma }^{\mathrm{\prime }}=0$, so constant curves are included. On a zero-dimensional manifold every smooth curve on an interval is locally constant and hence has zero velocity; the empty manifold has no such curves. A singleton parameter interval is excluded because [F2] supplies no derivative operator there. Included interval endpoints use the one-sided convention, and no point, chart, or curve is selected from a family, so no choice principle is used.

Metric compatibility and the symmetry of $g$ give $\frac{d}{dt}g({\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})=g(D_t{\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})+g({\gamma }^{\mathrm{\prime }},D_t{\gamma }^{\mathrm{\prime }})=2g(D_t{\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})=0$ by [F1] and [F2].

By [F3], $g({\gamma }^{\mathrm{\prime }},{\gamma }^{\mathrm{\prime }})$ is constant on the interval. It is nonnegative, so its nonnegative square root $\mathrm{\mid }{\gamma }^{\mathrm{\prime }}\mathrm{\mid }$ is constant as well. This includes the zero-speed constant geodesics and shows that a nonconstant geodesic never has zero velocity. In dimension zero the constant is zero; empty manifolds give no curves. Included parameter endpoints follow by continuity from the interior, and no choices are made.

Along the chart segment, ${\gamma }^{\mathrm{\prime }}={\dot{x}}^j{\mathrm{\partial }}_j$. The connection product rule and [F2] give $D_t{\gamma }^{\mathrm{\prime }}={\ddot{x}}^k{\mathrm{\partial }}_k+{\dot{x}}^j{\dot{x}}^i{\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_j=({\ddot{x}}^k+{\mathrm{\Gamma }}^k{}_{ij}(x){\dot{x}}^i{\dot{x}}^j){\mathrm{\partial }}_k\mathrm{.}$

The coordinate vectors are a basis at every point, so the vector in step 1.1 vanishes if and only if every displayed coefficient vanishes. By [F1], these two conditions are respectively equivalent to the intrinsic geodesic equation, proving both directions. For a constant curve all first and second derivatives vanish. In dimension zero both lists of equations are empty and both conditions hold; in dimension one the formula is $\ddot{x}+{\mathrm{\Gamma }}^1{}_{11}(x){\dot{x}}^2=0$. Included endpoints use one-sided derivatives, chart seams are handled on overlapping subintervals by the intrinsic equation, and no choices are made.

In the $2n$ coordinates $(x,v)$ supplied by [F2], the displayed expression has base components $v^i$ and fibre components $-{\mathrm{\Gamma }}^k{}_{ij}(x)v^i{v}^j$. The Christoffel functions are smooth, so every component is smooth. An integral curve of this local expression obeys exactly the first-order system in [F3].

At a zero vector $v=0$ both component lists vanish, so the zero section consists of stationary points of the local spray. For $n=1$ the formula is $v{\mathrm{\partial }}_x-{\mathrm{\Gamma }}^1{}_{11}(x)v^2{\mathrm{\partial }}_v$; for $n=0$ it is the zero vector field on the discrete zero section. Empty $M$ gives empty $TM$. No claim of overlap agreement is used here; that is the next lemma. The only choice principle is the explicitly assumed ${\mathrm{A}\mathrm{C}}_{\omega }$ in [F1]–[F2], used to obtain the global smooth-manifold structure on $TM$; the coordinate formula itself is choice-free.

On the overlap, $w^a=(\mathrm{\partial }{y}^a\mathrm{/}\mathrm{\partial }{x}^i)v^i$. Along a local integral curve of the $x$-formula, differentiation gives ${\dot{y}}^a=\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^i}{v}^i=w^a,$ and ${\dot{w}}^c=\frac{{\mathrm{\partial }}^2{y}^c}{\mathrm{\partial }{x}^i\mathrm{\partial }{x}^j}{v}^i{v}^j-\frac{\mathrm{\partial }{y}^c}{\mathrm{\partial }{x}^k}{\mathrm{\Gamma }}^k{}_{ij}{v}^i{v}^j\mathrm{.}$ Differentiating the inverse-coordinate identity twice gives $\frac{{\mathrm{\partial }}^2{y}^c}{\mathrm{\partial }{x}^i\mathrm{\partial }{x}^j}+\frac{\mathrm{\partial }{y}^c}{\mathrm{\partial }{x}^k}\frac{{\mathrm{\partial }}^2{x}^k}{\mathrm{\partial }{y}^a\mathrm{\partial }{y}^b}\frac{\mathrm{\partial }{y}^a}{\mathrm{\partial }{x}^i}\frac{\mathrm{\partial }{y}^b}{\mathrm{\partial }{x}^j}=0.$ Inserting this and [F2] yields ${\dot{w}}^c=-{\tilde{\mathrm{\Gamma }}}^c{}_{ab}{w}^a{w}^b$, exactly the $y$-formula.

Step 1.1 is the tangent-coordinate transformation law for the local vector fields, so the formulas glue to one vector field on $TM$. Their coordinate components are smooth by [F1], hence the glued field is smooth.

If $(x(t),v(t))$ is an integral curve, its first component equation says $v^i={\dot{x}}^i$ and its second says ${\ddot{x}}^k+{\mathrm{\Gamma }}^k{}_{ij}{\dot{x}}^i{\dot{x}}^j=0$; [F3] therefore makes $x(t)$ a geodesic and $(x,v)$ its velocity lift. Conversely, a geodesic and its velocity satisfy those two equations by [F3], so its lift is an integral curve. At $v=0$ the lift is stationary; dimensions zero and one reduce respectively to the empty system and the scalar calculation, and the empty bundle is harmless. Parameter endpoints are local and one-sided where included. No choice occurs in the overlap calculation; ${\mathrm{A}\mathrm{C}}_{\omega }$ remains the explicit hypothesis inherited from [F1].

Apply [F2] to the spray at the point $v\in TM$. It gives a unique maximal integral curve $Z_v:I_v\to TM$ on an open interval containing zero. By [F1], $Z_v$ is the velocity lift of ${\gamma }_{p,v}:=\pi \circ Z_v$. Since $Z_v(0)=v$, its base point is $p$, and the base component of the spray equation gives ${\gamma }_{p,v}^{\mathrm{\prime }}(0)=v$.

If a geodesic with these initial data existed on a larger interval, [F1] would make its velocity lift an integral curve of the spray extending $Z_v$, contrary to maximality. The same lift argument and integral-curve uniqueness prove uniqueness on every common interval. Thus $I_{p,v}=I_v$ and the geodesic is uniquely maximal.

By [F3], $\{(t,v):t\in I_v\}$ is open in $\mathbb{R}\times TM$ and $(t,v)\mapsto Z_v(t)$ is smooth. This is exactly $\mathcal{G}$ after writing $v$ together with its determined base point $p$. In induced tangent-bundle coordinates the projection $\pi (x,w)=x$ is smooth, so composing gives the asserted smooth geodesic evaluation.

For $v=0$, [F1] makes $Z_v$ stationary and the maximal geodesic is the constant curve on all of $\mathbb{R}$. In dimension zero every initial vector is zero; for empty $M$ there are no initial vectors. Each maximal domain is open, so it has no included finite endpoints. All conclusions concern one supplied initial vector at a time. The only choice principle is the declared ${\mathrm{A}\mathrm{C}}_{\omega }$, inherited exactly from the smooth-manifold structure on $TM$; [F2]–[F3] then apply without another family selection.

Put $\tilde{\gamma }=\gamma \circ \mathrm{\ell }$. The ordinary and covariant chain rules give ${\tilde{\gamma }}^{\mathrm{\prime }}(t)=a{\gamma }^{\mathrm{\prime }}(at+b)$ and $D_t{\tilde{\gamma }}^{\mathrm{\prime }}=a^2(D_s{\gamma }^{\mathrm{\prime }})(at+b)=0$ by [F1]. Hence $\tilde{\gamma }$ is geodesic.

By homogeneity of the norm in [F2], $\mathrm{\mid }{\tilde{\gamma }}^{\mathrm{\prime }}(t)\mathrm{\mid }=\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }{\gamma }^{\mathrm{\prime }}(at+b)\mathrm{\mid }$. If $a=0$, the reparametrized curve is constant and both formulas give zero; negative $a$ reverses the parameter and uses the absolute value. Dimensions zero and one require no change, empty manifolds have no curves, and included endpoints use the corresponding one-sided chain rule. The constants and curve are supplied, so no choice principle is used.

On $a^{-1}{I}_{p,v}$, the curve $\eta (t)={\gamma }_{p,v}(at)$ is geodesic by [F2], with $\eta (0)=p$ and ${\eta }^{\mathrm{\prime }}(0)=av$. The unique maximal solution in [F1] extends every solution with those initial data, so $a^{-1}{I}_{p,v}\subseteq I_{p,av}$ and $\eta (t)={\gamma }_{p,av}(t)$ throughout that domain.

If $a\mathrm{\ne }0$ and $I_{p,av}$ were larger than $a^{-1}{I}_{p,v}$, then $s\mapsto {\gamma }_{p,av}(s\mathrm{/}a)$ would extend ${\gamma }_{p,v}$ beyond $I_{p,v}$, contradicting maximality; applying the same argument in the other direction proves $I_{p,av}=a^{-1}{I}_{p,v}$. If $a=0$, both initial velocity and the right-hand curve are zero/constant, and [F1] gives the global domain $\mathbb{R}$. This also treats dimensions zero and one and all signs of $a$. The domains are open, so no finite endpoint is included. The only choice principle is the explicitly inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ for the geodesic-flow construction; no new selection is made.

The definition is intrinsic because [F1] makes $I_{p,v}$ unique. A geodesic remains in the connected component of its initial point, since the continuous image of its interval is connected; therefore requiring all $I_{p,v}=\mathbb{R}$ is equivalent to requiring the same condition separately on every component.

The zero vector gives the constant geodesic and already has domain $\mathbb{R}$. In dimension zero every vector is zero, so every boundaryless zero-manifold is geodesically complete; the empty manifold satisfies the universal condition vacuously. Failure means one explicitly existing initial vector has a finite end in its maximal open interval, so no included-endpoint ambiguity occurs. The only choice principle is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited from the construction of the maximal geodesics; the universal quantifier itself selects nothing.

Every tangent vector $v\in TM$ has the unique base point $p=\pi (v)$, and [F1] uniquely determines both $I_{p,v}$ and ${\gamma }_{p,v}$. Thus membership in $\mathcal{E}$ and the value ${\gamma }_{p,v}(1)$ are well-defined. The definition only evaluates curves whose maximal interval actually contains $1$ and therefore does not presume geodesic completeness.

The zero vector $0_p$ gives the constant geodesic on all of $\mathbb{R}$, so $0_p\in {\mathcal{E}}_p$ and ${\exp }_p(0_p)=p$. In dimension zero all tangent vectors are zero; if $M$ is empty, then $TM$ and $\mathcal{E}$ are empty and the displayed map is the unique empty function. Because $I_{p,v}$ is open, $1\in I_{p,v}$ is an interior-time condition rather than an included-endpoint convention; $\mathcal{E}$ may still be a proper subset of $TM$. The only choice principle used is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F1].

The map $j:TM\to \mathbb{R}\times TM$, $j(v)=(1,v)$, is smooth. By [F1] and [F2], $\mathcal{E}=j^{-1}(\mathcal{G})$, so $\mathcal{E}$ is open in $TM$. Every zero vector $0_p$ lies in $\mathcal{E}$ because its maximal geodesic is the constant curve on $\mathbb{R}$.

The restriction $j{\mathrm{\mid }}_{\mathcal{E}}:\mathcal{E}\to \mathcal{G}$ is smooth, and [F1] gives $\exp =G\circ j{\mathrm{\mid }}_{\mathcal{E}}$. Hence $\exp$ is smooth. For fixed $p$, ${\mathcal{E}}_p$ is the inverse image of the open set $\mathcal{E}$ under the smooth linear inclusion $T_{p}M\hookrightarrow TM$ and is therefore open in $T_{p}M$; the restriction ${\exp }_p$ is smooth.

If $M$ is empty, then $TM$, $\mathcal{E}$, and the zero section are empty, and openness and smoothness are vacuous. In dimension zero, $TM$ is the zero section and $\mathcal{E}=TM$; in dimension one the same slice and composition arguments apply unchanged. The zero-vector case was checked in step 1.1, and time $1$ is an interior point of each relevant open interval. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only through [F1] and [F2] to obtain the global smooth tangent-bundle/geodesic construction; taking a preimage and restricting a map require no further choice.

Suppose first that $t\mathrm{\ne }0$. By [F1], $1\in I_{p,tv}$ iff $t\in I_{p,v}$, and on that domain ${\gamma }_{p,tv}(1)={\gamma }_{p,v}(t)$. Applying [F2] proves both the domain equivalence and the exponential identity.

If $t=0$, then $0\in I_{p,v}$, while [F1] makes ${\gamma }_{p,0}$ the constant curve on $\mathbb{R}$; hence $0v\in {\mathcal{E}}_p$ and ${\exp }_p(0)=p={\gamma }_{p,v}(0)$. Thus the equivalence and identity also hold at zero.

Let $v\in {\mathcal{E}}_p$ and $s\in [0,1]$. Then $1\in I_{p,v}$ by [F2]. Since $I_{p,v}$ is an interval containing $0$, it contains $s$, so step 1.1 or 1.2 gives $sv\in {\mathcal{E}}_p$. Thus every fibre domain is star-shaped about zero. Negative parameters are covered by step 1.1 whenever the corresponding geodesic time lies in the maximal interval.

On an empty manifold there are no initial vectors. In dimension zero only step 1.2 occurs, and in dimension one the proof is unchanged. Zero velocity and zero scale were handled explicitly; membership is always at the interior time $1$ of an open maximal interval, including when the original time is a finite endpoint candidate. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]--[F2], and no additional selection is made.

Fix $w\in T_{p}M$. By [F1], the straight line $c(s)=sw$ lies in ${\mathcal{E}}_p$ for all sufficiently small $s$, and $c^{\mathrm{\prime }}(0)$ corresponds to $w$ under $T_{0_p}(T_{p}M)\cong T_{p}M$. The curve definition of the differential and [F2] give $d({\exp }_p{)}_{0_p}(w)={\frac{d}{ds}\mid }_0{\exp }_p(sw)={\frac{d}{ds}\mid }_0{\gamma }_{p,w}(s)=w,$ because ${\gamma }_{p,w}^{\mathrm{\prime }}(0)=w$. Hence the differential is the identity.

For $w=0_p$, both sides in step 1.1 are zero. In dimension zero the identity is the unique map on the zero vector space; dimension one is the same one-vector computation. If $M$ is empty there is no point $p$, so the statement is vacuous. Only an arbitrarily small open parameter interval around zero is used, not an endpoint of the exponential domain. The stated ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through [F1]--[F2], and differentiating a fixed curve introduces no choice.

Take $R,q,C$ from [F1] and choose the explicit positive number $\delta =(1-q)R\mathrm{/}(2C)$. Put $W=B(b,\delta )$. For $y\in W$ and $x\in \bar{B}(a,R)$, $\mathrm{\parallel }{T}_y(x)-a{\mathrm{\parallel }}_2\le \mathrm{\parallel }{T}_y(x)-T_y(a){\mathrm{\parallel }}_2+\mathrm{\parallel }B(y-b){\mathrm{\parallel }}_2\le qR+C\delta =(1+q)R\mathrm{/}2<R\mathrm{.}$ Thus $T_y$ maps the closed ball strictly into its open interior.

The closed ball is complete without choice. Indeed, a Cauchy sequence $(x_k)$ in it is Cauchy in ${\mathbb{R}}^n$, so [F2] gives its unique limit $x$. The triangle inequality yields $\mathrm{\parallel }x-a{\mathrm{\parallel }}_2\le \mathrm{\parallel }x-x_k{\mathrm{\parallel }}_2+R$ for every $k$; if $\mathrm{\parallel }x-a{\mathrm{\parallel }}_2>R$, choosing $k$ with $\mathrm{\parallel }x-x_k{\mathrm{\parallel }}_2<\mathrm{\parallel }x-a{\mathrm{\parallel }}_2-R$ is a contradiction. Hence $x$ remains in the ball. The ball is nonempty because it contains $a$.

For each fixed $y\in W$, [F1], step 1.1, step 1.2, and [F2] give a unique fixed point $g(y)\in \bar{B}(a,R)$. This defines a function without a choice axiom: $g(y)$ is the unique object satisfying the displayed fixed-point property. Its equation is $B(y-f(g(y)))=0$, hence $f(g(y))=y$ because $B$ is injective; step 1.1 puts $g(y)$ in $B(a,R)$.

If $x,z\in \bar{B}(a,R)$ and $f(x)=f(z)=y$, then $T_y(x)=x$ and $T_y(z)=z$, so [F1] gives $\mathrm{\parallel }x-z{\mathrm{\parallel }}_2\le q\mathrm{\parallel }x-z{\mathrm{\parallel }}_2$ and therefore $x=z$. Define $V=B(a,R)\cap f^{-1}[W]$. By [F3], $V$ is open; it contains $a$, lies in $U$, and step 2.1 together with injectivity shows that $f{\mathrm{\mid }}_V:V\to W$ is bijective with inverse $g$.

For $y,z\in W$, the fixed-point equations and [F1] give $\mathrm{\parallel }g(y)-g(z){\mathrm{\parallel }}_2\le q\mathrm{\parallel }g(y)-g(z){\mathrm{\parallel }}_2+C\mathrm{\parallel }y-z{\mathrm{\parallel }}_2,$ so $\mathrm{\parallel }g(y)-g(z){\mathrm{\parallel }}_2\le C(1-q{)}^{-1}\mathrm{\parallel }y-z{\mathrm{\parallel }}_2$. Thus $g$ is Lipschitz and continuous.

Fix $y\in W$, put $x=g(y)$ and $L=Df(x)$. For small $h$ with $y+h\in W$, put $k=g(y+h)-x$. Step 3.2 gives $\mathrm{\parallel }k{\mathrm{\parallel }}_2=O(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$, while [F3] and $f(g(y+h))-f(g(y))=h$ give $h=Lk+r(k)$ with $\mathrm{\parallel }r(k){\mathrm{\parallel }}_2=o(\mathrm{\parallel }k{\mathrm{\parallel }}_2)$. The inverse bound in [F1] therefore gives $k=L^{-1}h-L^{-1}r(k)=L^{-1}h+o(\mathrm{\parallel }h{\mathrm{\parallel }}_2)$, including the case $k=0$. Hence $Dg(y)=Df(g(y){)}^{-1}$. The chain rule in [F5] also gives $Df(g(y))Dg(y)=I$ from $f\circ g={\mathrm{id}}_W$, consistently with this formula.

The map $g$ is $C^1$: it is continuous by step 3.2, $Df\circ g$ is continuous, and [F4] makes its inverse matrix $Dg$ continuous. Inductively, suppose $g$ is $C^r$ for some $r\ge 1$. Because $f$ is smooth, the matrix entries of $Df$ are $C^r$; [F4] makes $Df\circ g$ and then $(Df\circ g{)}^{-1}=Dg$ of class $C^r$. Thus the first partial derivatives of $g$ are $C^r$, so [F4] makes $g$ of class $C^{r+1}$. Induction proves $g$ is smooth, and [F4] and step 3.1 make $f{\mathrm{\mid }}_V$ a diffeomorphism.

The hypothesis $n\ge 1$ excludes the zero-dimensional Euclidean convention; dimension one is included verbatim. The datum $a\in U$ makes the empty-domain case impossible. Invertibility excludes a degenerate derivative, while zero increments are covered in step 4.1. All domains are open, and the closed ball is used only as the complete space for iteration, so no boundary point is asserted to lie in $V$. The construction chooses explicit $\delta$, starts every Newton iteration at the specified point $a$, and defines each value by uniqueness; finite induction on derivative order and unique Euclidean limits use no choice axiom.

Suppose $n\ge 1$. Choose one smooth chart $x:Q\to x(Q)\subseteq {\mathbb{R}}^n$ at $p$ and let $L:{\mathbb{R}}^n\to T_{p}M$ be the inverse of the chart-induced linear isomorphism $d{x}_p$. The local representative $F=x\circ {\exp }_p\circ L$ is defined and smooth on the open neighbourhood $L^{-1}({\mathcal{E}}_p\cap {\exp }_p^{-1}(Q))$ of $0$ by [F1]--[F2], satisfies $F(0)=x(p)$, and has derivative $DF(0)=d{x}_p\circ I\circ L=I$ by [F1]--[F2].

Apply [F3] to $F$. It yields open neighbourhoods $P$ of $0$ and $W$ of $x(p)$ on which $F$ is a diffeomorphism. Since $P$ is open about $0$, choose $r>0$ with $B(0,r)\subseteq P$. Put ${\tilde{U}}_p=L(B(0,r))$. This set is open and star-shaped about $0_p$, and the restriction of ${\exp }_p$ is a diffeomorphism onto its image: it is the chart conjugate of the restriction of $F{\mathrm{\mid }}_P$, whose inverse remains smooth. Its image $U_p$ is open because $F(B(0,r))$ is open in $W$ under the homeomorphism $F{\mathrm{\mid }}_P$, and $x^{-1}$ is a chart homeomorphism. Finally $0_p\in {\tilde{U}}_p$ and ${\exp }_p(0_p)=p$, so $p\in U_p$.

If $n=0$, a manifold chart shows that $\{p\}$ is open and $T_{p}M=\{0_p\}$. Take ${\tilde{U}}_p=\{0_p\}$ and $U_p=\{p\}$; the exponential is the unique bijection and both it and its inverse are smooth under the zero-dimensional convention. Dimension one is included in steps 1.1--2.1. An empty manifold has no $p$, so the universal statement is vacuous. The source ball is open, so no sphere endpoint is included, and it contains the degenerate zero vector. ${\mathrm{A}\mathrm{C}}_{\omega }$ is used only through [F1]; the choice-free inverse theorem [F3] and choosing one chart and one positive radius for the fixed $p$ require no family choice.

Because ${\exp }_p{\mathrm{\mid }}_D$ is a diffeomorphism, its inverse is a well-defined smooth map $U\to D$. The supplied basis makes $E_e$ a specified linear isomorphism, so $E_e^{-1}(D)$ is open and $x_e$ is a diffeomorphism onto it. Thus the intrinsic normal neighbourhood does not depend on coordinates, while the displayed coordinate list records exactly its dependence on the supplied basis.

Star-shaped means that $v\in D$ implies $tv\in D$ for every $0\le t\le 1$, including both endpoints. The zero vector belongs to $D$ and maps to $p$. In dimension zero, the ordered basis is the empty tuple, ${\mathbb{R}}^0$, $T_{p}M$, $D$, and $U$ are singletons, and the formula gives the unique chart; dimension one is literal. If $M$ is empty there is no centre $p$. The basis is supplied rather than chosen, so the only choice principle is the stated ${\mathrm{A}\mathrm{C}}_{\omega }$ inherited through [F1].