= Solution
For $v\in T_pS$ sufficiently small, let $\gamma_v$ be the <geodesic> with $\gamma_v(0)=p$, $\dot\gamma_v(0)=v$, and define $\exp_p(v)=\gamma_v(1)$. The domain consists more generally of initial velocities whose <geodesic> exists through time one. Uniqueness and affine rescaling give $\exp_p(tv)=\gamma_v(t)$ near zero, so
$$
\boxed{d\exp_p|_0(v)=v.}
$$
Thus the differential is the identity under the natural identification of tangent spaces, and the <inverse function theorem> makes the <exponential map> a <local diffeomorphism> near zero.
A punctured plane has <geodesics> reaching its missing point in finite time, so its <exponential map> need not be global; adding the point remedies this example. But \b[an extension cannot always remedy the failure]. On the <punctured circular cone> $S=\{(r\cos\theta,r\sin\theta,r):r>0\}$, radial generators are <geodesics> and reach the missing point in finite intrinsic length. Any globally defined extension would have to include their limiting point $(0,0,0)$. The cone's tangent planes have different limits as $\theta$ varies; no smooth embedded surface through that point can contain this <punctured circular cone>. Hence no such extension exists for this $S$.
For an <oriented surface> and a closed topological disc $D$ with smooth positively oriented boundary, the <Gauss-Bonnet theorem> with boundary says
$$
\boxed{\int_D K\,dA+\int_{\partial D}k_g\,ds=2\pi\chi(D)=2\pi.}
$$
Here $k_g$ is signed <geodesic curvature> with the disc on the left. A smooth boundary has no corner terms; piecewise smooth boundaries require exterior-angle terms.
For the flat disc in the problem this gives $\int_{\partial D}k_g\,ds=2\pi$. Conditions (i) and (ii) mean that the proposed replacement disc has the same boundary and agrees with the original <smooth surface> from the outside. Smoothness therefore gives the same tangent planes, metric <derivatives> and boundary <geodesic curvature> on the replacement, with consistent <orientation>. Applying Gauss-Bonnet to that disc forces
$$
\boxed{\int_{\widetilde D}\widetilde K\,d\widetilde A=0.}
$$
Thus these gluing conditions fix the <total curvature> of any smooth replacement disc, regardless of its detailed interior shape.
The nonnegative continuous <Gaussian curvature> on the compact replacement disc, together with its zero integral established by Gauss-Bonnet, forces it to vanish at every point of that disc. Outside it the replacement surface agrees with the original flat surface, so its curvature also vanishes there; smoothness covers the common boundary. Consequently no point of the replacement surface can have positive <Gaussian curvature>. Therefore \b[no surface can satisfy all three conditions]. A disc with some positive curvature could only have compensating negative curvature or alter the smooth boundary gluing, both excluded here.
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