# dφ is bijective but φ is not a lie group isomorphism

suppose G and H connected Lie groups. Is there $$\phi: G \to H$$ a morphism of Lie groups such that $$d\phi$$ is bijective but $$\phi$$ is not an isomorphism of Lie groups?

I know that $$d\phi$$ surjective and H connected $$\Rightarrow \phi$$ surjective. And I know that $$d\phi$$ injective $$\Rightarrow Ker \phi$$ is discrete.

So to come up with a potential example, I need to prove that $$Ker\phi$$ could be non trivial. Is that a possibility?

Many thanks for your hints or help.

Yes. Let $$\phi: \mathbb{R} \to \frac{\mathbb{R}}{\mathbb{Z}}$$ be the natural map. Then $$d\phi: \mathbb{R} \to \mathbb{R}$$ is the identity.
• Thank you! Please correct me if I am wrong: $d\phi: Lie(\Bbb R) \to Lie(\Bbb R/\Bbb Z)$. I understand that $Lie(\Bbb R)\cong T\Bbb R = \Bbb R$ but is $Lie(\Bbb R/\Bbb Z) \cong T(\Bbb R/\Bbb Z) = \Bbb R$? – PerelMan Apr 9 at 14:20
• @PerelMan yes. The metaphor is that the tangent space at the identity is located infinitesimally at zero so it can't "see" that you've killed $1$ which is far away, so the tangent spaces are just the same. More rigorously, if you think of a tangent vector as something that eats germs of smooth functions, a smooth function in a neighborhood of $0$ on $\mathbb{R}$ is the same thing as a smooth function in a neighborhood of $0$ on $\mathbb{R}/\mathbb{Z}$ once the neighborhood gets small enough. – hunter Apr 9 at 14:26