Geometrically speaking, one might consider the hyperbolic space $H^2$ and the sphere $S^2$ as manifolds which deviate from the euclidean plane $\mathbb{R}^2$ in exactly opposite ways – $H^2$ having constant negative curvature, $S^2$ constant positive curvature.
However, $S^2$ also has a quite different topology from $\mathbb{R}^2$, whereas $H^2$ is homeomorphic to it. What's the reason for this qualitative difference, or rather for the topological non-difference between $H^2$ and $\mathbb{R}^2$?
Compute the length of a perimeter of a circle in all the geometries. It is $2\pi\sin(r) = \Theta(1)$ in spherical geometry, $2\pi r = \Theta(r)$ in Euclidean geometry, and $2\pi\sinh(r) = \Theta(e^r)$ in hyperbolic geometry.
There is a significant difference between all the three situations -- in Euclidean perimeters and areas grow polynomially with the radius, while in hyperbolic geometry they grow exponentially. In spherical geometry everything is bounded. However, topology cares only about whether it is bounded or not, so Euclidean and hyperbolic are the same topologically.
Not sure whether it helps other people, but I like to imagine the discrete version of this: build a graph out of equiltateral triangles, where $k$ meet in a vertex. For $k<6$ you get a finite spherical graph (icosahedron for $k=5$), for $k=6$ you get an infinite triangular grid on the plane, and for $k>6$ you get a tesselation of a hyperbolic plane. Curvature corresponds to $6-k$. Again, the number of vertices in distance $d$ is bounded for $k<6$, polynomial for $k=6$, and exponential for $k>6$.