Start with the Euclidean space and take the smooth function defined by . In components, this is , where the are the canonical coordinates on . We can easily calculate the derivative in these coordinates: . This is the zero function if and only if , and so has rank at any nonzero point . The point is a critical point, and every other point is regular.
On the image side, we see that , so the only critical value is . Every other value is regular, though is empty for . For we have a nonempty preimage, which by our result is a manifold of dimension . This is the -dimensional sphere of radius , though we aren’t going to care so much about the radius for now.
Anyway, is this really the same sphere as before? Remember, when we first saw the two-dimensional sphere as an example, we picked coordinate patches by hand. Now we have the same set of points — those with a fixed squared-distance from the origin — but we might have a different differentiable manifold structure. But if we can show that the inclusion mapping that takes each of our handcrafted coordinate patches into is an immersion, then they must be compatible with the submanifold structure.
We only really need to check this for a single patch, since all six are very similar. We take the local coordinates from our patch and the canonical coordinates on to write out the inclusion map:
Then we use these coordinates to calculate the derivative
This clearly always has rank for , and so the inclusion of our original sphere into is an immersion, which must then be equivalent to the inclusion of the submanifold , since they give the same subspace of .