# Is Pull back of Lie group a Lie group?

Suppose $$f:M\rightarrow N$$ is a submersion. Then, $$M\times_NM$$ is a smooth manifold from Transversal theorem.

Suppose $$\theta:G\rightarrow H$$ is a morphism of Lie groups. Assume that it is a submersion. Does it imply $$G\times_H G$$ is a Lie group?

As $$G\rightarrow H$$ is submersion, $$G\times_H G$$ is a smooth manifold. Does it have to be Lie group?

When will $$G\times_H G$$ is a Lie group?

Yes it’s always a Lie group.

Closure under multiplication and inverses you can check, so it’s a group. For smoothness, note $$M\times_NM$$ isn’t just a manifold, but an embedded submanifold of $$M\times M$$. The multiplication and inversion for $$G\times_HG$$ are restricted from $$G\times G$$ to the submanifold, hence smooth.

• Ok. Restriction of smooth map to embeded submanifold is smooth.. So, it i the case.. Thanks
– user537667
Jan 20, 2019 at 5:35
• @PraphullaKoushik No problem!
– Ben
Jan 20, 2019 at 6:00
• I have added more details as a community wiki answer. Let me know if I am missing something. Thank you :)
– user537667
Jan 20, 2019 at 9:21