If we have a smooth map and a regular value of , we know that the preimage is a smooth -dimensional submanifold. It turns out that we also have a nice decomposition of the tangent space for every point .
The key observation is that the inclusion induces an inclusion of each tangent space by using the derivative . The directions in this subspace are those “tangent to” the submanifold , and so these are the directions in which doesn’t change, “to first order”. Heuristically, in any direction tangent to we can set up a curve with that tangent vector which lies entirely within . Along this curve, the value of is constantly , and so the derivative of is zero. Since the derivative of in the direction only depends on and not the specific choice of curve , we conclude that should be zero.
This still feels a little handwavy. To be more precise, if and is a smooth function on a neighborhood of , then we calculate
since any tangent vector applied to a constant function is automatically zero. Thus we conclude that . In fact, we can say more. The rank-nullity theorem tells us that the dimension of and the dimension of add up to the dimension of , which of course is . But the assumption that is a regular point means that the rank of is , so the dimension of the kernel is . And this is exactly the dimension of , and thus of its tangent space ! Since the subspace has the same dimesion as , we conclude that they are in fact equal.
What does this mean? It tells us that not only are the tangent directions to contained in the kernel of the derivative , every vector in the kernel is tangent to . Thus we can break down any tangent vector in into a part that goes “along” and a part that goes across it. Unfortunately, this isn’t really canonical, since we don’t have a specific complementary subspace to in mind. Still, it’s a useful framework to keep in mind, reinforcing the idea that near the subspace the manifold “looks like” the product of (from ) and , and we can even pick coordinates that reflect this “decomposition”.