The "rank" of f:A-> B is the dimension of f(A) as a subspace of B. Since the rank of g:B-> C is the same as the dimension of C, g(B) is all of C and, in particular, g(f(A)) has the same dimension as f(A).
Let and be two differentiable functions and , and be manifolds. If the rank of at is equal to the dimension of , show that and have the same rank at .
Something tells me that I have to use the fundemental equation of dimension, but I can't work it out.