Overlap structure of arc-length parametrizations of a 1-manifold

On $S=f^{-1}(h(J))$ put $\phi =h^{-1}\circ f$. Differentiating $f=h\circ \phi$ and taking lengths gives $\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }=1$. On each component $D$ of $S$, continuity makes ${\phi }^{\mathrm{\prime }}$ constant, so $\phi (s)={\epsilon }_{D}s+c_D$, with ${\epsilon }_D=\pm 1$. The graph is closed in $I\times J$: it is the inverse image of the diagonal of the Hausdorff manifold $M$ under $(f,h)$. Its segments are maximal intersections of their affine lines with $I\times J$, since closedness and the local diffeomorphism property prevent a segment from stopping where both coordinates remain in the relative interiors of their intervals. Included interval endpoints are retained in this assertion.

Each end of a segment therefore reaches an end of $I$ or $J$. At most one segment can reach any one of the four sides: two reaching an $I$-side would have overlapping $s$-projections, contradicting single-valuedness, and two reaching a $J$-side would have overlapping $t$-projections, contradicting injectivity. This includes unbounded ends, since two tails towards the same infinite end overlap. Every segment consumes two distinct sides, so there are at most two segments. With two segments their projections are disjoint on both axes; they must occupy opposite corners, joining left to top and bottom to right (slope $+1$), or left to bottom and top to right (slope $-1$). Thus their slopes agree.

If there is one component, extend its affine expression to $L:\mathbb{R}\to \mathbb{R}$. Maximality of the segment gives $S=I\cap L^{-1}(J)$. The union $I\cup L^{-1}(J)$ is an interval, and $f$ and $h\circ L$ agree on the overlap. Their glued map is smooth and unit-speed. If $f(s)=h(L(s^{\mathrm{\prime }}))$, then $s\in S$ and injectivity of $h$ gives $L(s)=L(s^{\mathrm{\prime }})$, hence $s=s^{\mathrm{\prime }}$. On each open domain piece it is the given local diffeomorphism $f$ or $h\circ L$, so the glued map is a diffeomorphism onto the open union, as required.

In the two-component case reflect a parameter if needed so both slopes are $+1$. The opposite-corner arrangement of 2.1 has $I=(a,d)$, transition expressions $s+p$ on $(a,b)$ and $s+q$ on $(c,d)$, and $J=(c+q,b+p)$, where $a<b\le c<d$ and $d+q\le a+p$. These endpoints are finite: $b,c$ lie inside $I$, while $a+p,d+q$ lie inside $J$. The endpoints of $I,J$ are excluded, since inclusion of one would equate a boundary point of one parametrization with an interior point of the other, by continuity of the transition and preservation of boundary under diffeomorphisms. Put $L=p-q>0$. On $\mathbb{R}\mathrm{/}L\mathbb{Z}$ define $H([t])=f(a+t)$ for $0<t<d-a$ and $H([t])=h(a+q+t)$ for $d-a\le t\le L$, identifying $L$ with $0$. These cover the circle because $d-a\le L$. At $t=d-a$ and $t=0$ the adjacent formulas agree in the $h$-chart with the same affine coordinate, so $H$ is smooth and unit-speed across both seams.

The first branch of $H$ parametrizes $f(I)$ injectively; the remaining arc parametrizes the part of $h(J)$ outside $f(I)$, including the two seams. The only identifications are the stated transition relations, so $H$ is injective and its image is $f(I)\cup h(J)$. It is a local diffeomorphism, hence has open image; its compact image is closed in the Hausdorff $M$. Connectedness forces its image to be all of $M$, and the bijective local diffeomorphism $H$ proves $M\cong S^1$. The given metric and parametrizations require no choice principle.