Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

If $\phi:M\longrightarrow N$ is an injective smooth map between two manifolds, then is $d\phi_m:M_m\longrightarrow N_{\phi(m)}$, the induced map between the tangent spaces injective too?

I tried the following : If $v\in M_m$ is such that $d\phi_m(v)=0$, then for all $g$, $C^{\infty}$ function in a neighbourhood of $\phi(m)$, $d\phi_m(v)(g)=0$, that is $v(g\circ\phi)=0$ for all such $g$. From this can we conclude that $v$ is the $0$ tangent vector.

I got this doubt when I was trying to understand the definition of an immersion. I was wondering if $\phi$ being injective will automatically make it an immersion.

share|cite|improve this question
up vote 6 down vote accepted

The answer is no. Let $\phi:\mathbb R\rightarrow\mathbb R^2$ be $\phi(t)=(t^3,t^9)$. Then the derivative at $t=0$ is not injective.

share|cite|improve this answer
You can also use the function $f(x)=x^3$ from the real line to itself. – studiosus Feb 11 '14 at 10:33

If the map$\phi:M\longrightarrow N$ is an diffeomorphism then $d\phi_m:M_m\longrightarrow N_{\phi(m)}$ is an isomorphism

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.