Compact geodesic locally CAT(1) spaces are CAT(1) exactly when they contain no short circle

A short circle obstructs CAT(1). An isometric copy of $S_{\mathrm{\ell }}^1$, $\mathrm{\ell }<2\pi$, has geodesic triangle sides that remain geodesics in $X$, with exactly the same side and cross distances. The circle's failed CAT(1) test in [F1] is therefore a failed test in $X$. This proves the forward implication of (i). Empty and one-point spaces are CAT(1) and contain no circle; henceforth the non-CAT(1) case is nonempty.

A positive uniform uniqueness radius. Consider the family of all CAT(1) closed balls with radii $0<\rho <\pi \mathrm{/}4$; their open half-radius balls cover $X$. Shrinking is legitimate: balls of radius $<\pi \mathrm{/}2$ in a CAT(1) chart are convex, since each center-and-endpoints triangle has perimeter $<4\rho <2\pi$ and its spherical comparison stays in the model ball. The open half-radius balls cover $X$; take finitely many, with radii ${\rho }_i$, and put $\epsilon ={\min }_i{\rho }_i\mathrm{/}4$. If $d(x,y)<\epsilon$ and $x$ lies in the $i$th half-ball, every geodesic from $x$ to $y$ lies in its full ball, since each of its points is within $d(x,y)$ of $x$. Any two such geodesics coincide by the CAT(1) test on their digon with a zero third side, whose perimeter is $2d(x,y)<2\pi$. Thus $r:=\mathrm{injrad}(X)\ge \epsilon >0$.

Continuity under short uniqueness. Suppose $0<R\le \pi$ and all pairs at distance $<R$ have unique geodesics. If endpoints $u_n,v_n$ converge to $u,v$ with $d(u,v)<R$, parametrize their geodesics on $[0,1]$. Their speeds $d(u_n,v_n)$ are bounded, hence they are equicontinuous into compact $X$. By [F2] every subsequence has a uniformly convergent further subsequence; the distance equality $d(c_n(s),c_n(t))=\mathrm{\mid }s-t\mathrm{\mid }d(u_n,v_n)$ passes to the limit by [F4], making that limit the unique geodesic from $u$ to $v$. The entire sequence converges uniformly: failure would give a subsequence uniformly separated from that geodesic and contradict its convergent further subsequence. This proves endpoint continuity in the uniform metric. AC is used through the Ascoli corollary.

Angle comparison below the uniqueness threshold. For a triangle of perimeter $P<2R$, every side has length $<R$. For any point $v$ of the side opposite $p$, the two boundary routes give $2d(p,v)\le P<2R$. Step 1.3 supplies a continuous sweep of the unique geodesics $[p,v]$. If $p$ is off that side, patchwork [F3] gives angle comparison. If $p$ lies on that side, uniqueness makes all sides the corresponding subsegments and the triangle is isometric to its collinear comparison. Thus every triangle of perimeter $<2R$ has vertex angle comparison.

Comparison at scale $R$ and the compact uniqueness criterion. Under the hypothesis of (iii), use step 2.1 with that $R$. For a point $v$ interior to the side $[q_0,q_1]$ of a triangle of perimeter $P<2R$, the two triangles $(p,q_0,v)$ and $(p,v,q_1)$ have perimeter at most $P$. Their actual angles at $v$ have sum at least $\pi$ by [F3], and therefore so do their comparison angles. Glue their models along the common side $[\bar{p},\bar{v}]$ on opposite sides; the four boundary lengths sum to $P<2R\le 2\pi$. Alexandrov's marked-point comparison [F3] then gives $d(p,v)\le d_S(\bar{p},\bar{v})$ in the comparison triangle of $(p,q_0,q_1)$. Collinear submodels follow by the closed limiting cosine inequalities; $p=v$ or $p$ on the opposite side is already trivial by uniqueness. This proves every vertex-to-opposite-side inequality. It implies all-pair comparison as follows. For $x\in [p,q]$, $y\in [p,z]$, first apply it in $(p,x,z)$ to obtain $d(x,y)\le d_S(\widehat{x},\widehat{y})$; then apply it in $(p,q,z)$ to obtain $d(x,z)\le d_S(\bar{x},\bar{z})$. These subtriangles have perimeter at most the original perimeter. The spherical cosine rule shows that the second inequality bounds the model angle at $p$ in $(p,x,z)$ by the original model angle at $\bar{p}$, and the first then bounds $d(x,y)$ by $d_S(\bar{x},\bar{y})$: for fixed two positive sides $<\pi$, the opposite side increases with the included angle because both sines are positive. Zero sides and coinciding points give equality directly. This proves (iii); the sweep, both split triangles and both all-pair subtriangles stay within the same perimeter bound $2R$. Taking $R=\pi$ proves that short uniqueness makes $X$ CAT(1). Conversely CAT(1) implies uniqueness below $\pi$ by its digon test. We have proved, rather than assumed, the compact short-uniqueness criterion.

The first failure is below $\pi$. Suppose $X$ is not CAT(1). Step 3.1 gives two distinct geodesics between points of distance $L<\pi$. Uniqueness cannot hold at any radius greater than $L$, so $0<r\le L<\pi$. For every pair at distance $<r$, uniqueness holds: choose a radius in the defining set strictly greater than that distance, using the definition of supremum. In particular steps 1.3 and 2.1 apply with $R=r$.

A limiting minimizing digon. For each positive integer $n$ choose a pair of distinct geodesics with common endpoints and length $L_n<r+{\delta }_n$, where ${\delta }_n>0$ tends to zero and $r+{\delta }_n<\pi$; such a pair exists by the definition of $r$, and $L_n\ge r$. This countable selection uses AC. Parametrize both sides on $[0,1]$; they are uniformly Lipschitz with speeds $L_n$. Apply [F2] to the first sides and then to the corresponding second sides to obtain simultaneous uniform limits $c,c^{\mathrm{\prime }}$ and limits $x,y$ of the endpoints. Passing the distance equalities to the limit shows that both are geodesics from $x$ to $y$ of length $r$.

Explicit noncollapse of the digons. Suppose $c=c^{\mathrm{\prime }}$. Their midpoints $m_n,m_n^{\mathrm{\prime }}$ then have distance $h_n\to 0$. Put $a_n=L_n\mathrm{/}2<\pi \mathrm{/}2$. For large $n$, $L_n+h_n<2r$; thus each triangle $(x_n,m_n,m_n^{\mathrm{\prime }})$ and $(y_n,m_n,m_n^{\mathrm{\prime }})$ has perimeter $2a_n+h_n<2r$ and angle comparison by steps 2.1 and 4.1. If $h_n>0$, its comparison angle ${\theta }_n$ at $m_n^{\mathrm{\prime }}$ satisfies $\cos {\theta }_n=\cos a_n(1-\cos h_n)\mathrm{/}(\sin a_n\sin h_n)>0$, so ${\theta }_n<\pi \mathrm{/}2$. But the two actual angles at $m_n^{\mathrm{\prime }}$ sum to at least $\pi$, because $m_n^{\mathrm{\prime }}$ is interior to a geodesic and [F3] applies; comparison bounds their sum by $2{\theta }_n<\pi$, a contradiction. Consequently $h_n=0$. Since $a_n<r$ for large $n$, uniqueness identifies both halves through the common midpoint, contradicting the distinctness of the original digon sides. Therefore $c\mathrm{\ne }{c}^{\mathrm{\prime }}$.

Opposite points have distance $r$. Use arclength parameters on $c,c^{\mathrm{\prime }}$ and take $z=c(a)$, $z^{\mathrm{\prime }}=c^{\mathrm{\prime }}(b)$ with $a+b=r$. If $a=0$ or $b=0$ their distance is $r$. Otherwise $a,b>0$ and $h=d(z,z^{\mathrm{\prime }})\le r$ by the two routes through $x,y$. Suppose $h<r$. If $h=0$, the endpoint segments all have lengths $a,b<r$ and uniqueness makes both digon sides coincide, a contradiction. If $h>0$, both triangles $(x,z,z^{\mathrm{\prime }})$, $(y,z,z^{\mathrm{\prime }})$ have perimeter $r+h<2r$, hence angle comparison. Their model angles ${\theta }_x,{\theta }_y$ at $z^{\mathrm{\prime }}$ satisfy $\cos {\theta }_x+\cos {\theta }_y=\frac{\sin r(\cos (a-b)-\cos h)}{\sin a\sin b\sin h}$. The reverse triangle inequality gives $h\ge \mathrm{\mid }a-b\mathrm{\mid }$. If $h>\mathrm{\mid }a-b\mathrm{\mid }$, the displayed quantity is positive, whereas ${\theta }_x+{\theta }_y\ge \pi$ (the actual angle sum is at least $\pi$) would give $\cos {\theta }_x+\cos {\theta }_y=2\cos (({\theta }_x+{\theta }_y)\mathrm{/}2)\cos (({\theta }_x-{\theta }_y)\mathrm{/}2)\le 0$. Thus $h=\mathrm{\mid }a-b\mathrm{\mid }$. If $a\ge b$, the equality $d(x,z)=d(x,z^{\mathrm{\prime }})+d(z^{\mathrm{\prime }},z)$ puts $z^{\mathrm{\prime }}$ on a geodesic from $x$ to $z$; that geodesic is unique since $a<r$, so $z^{\mathrm{\prime }}$ lies on $c$. The unique segments from $x$ and $y$ to $z^{\mathrm{\prime }}$ (of lengths $b,a<r$) are then the corresponding subsegments of both sides, making $c=c^{\mathrm{\prime }}$. If $b\ge a$, interchange the sides. This contradiction proves $h=r$.

All circle distances, and conclusion. For arbitrary $c(s),c^{\mathrm{\prime }}(t)$ let $w=c^{\mathrm{\prime }}(r-s)$ be opposite $c(s)$. Step 7.1 and the reverse triangle inequality give $d(c(s),c^{\mathrm{\prime }}(t))\ge r-\mathrm{\mid }t-(r-s)\mathrm{\mid }=\min \{s+t,2r-s-t\}$; the two routes through $x,y$ give the reverse bound. Distances on either single side already equal parameter differences. Thus traversing $c$ and then $c^{\mathrm{\prime }}$ backwards gives an isometry $S_{2r}^1\to c([0,r])\cup c^{\mathrm{\prime }}([0,r])$: the distance formula also excludes every additional identification. Since $0<r<\pi$, this proves (ii); together with step 1.1 it proves (i).