Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $D\subset \mathbb R^2$, open set. We have a smooth map $f: D\times (0,1)\to \mathbb R^3$. There is a smooth map $u$ from some open subset of $\mathbb R^3$ to $ D$ such that $$u(f(x,y,t)) = (x,y)$$ Show that $u_*$ derivative of $u$ is of rank $2$. $t$ is parameter for $(0,1)$ direction and $(x,y)\in D$.

Any suggestion comment please.

share|cite|improve this question
up vote 2 down vote accepted

chain rule shows $u_* f_* = \operatorname {id}_* = \operatorname {id}$ which means that $u_*$ is surjective, i.e. has full rank

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.