0

I am trying to understand the following problem:

Given smooth injective Lie group homomorphism $i_j: H_i \to G$, $j\in \{1,2\}$, such that $i_1(H_1) = i_2(H_2)$, there exists an unique map $\varphi: H_1 \to H_2$ such that $i_1 = i_2\circ \varphi$, and which is an isomorphism of groups.

I'm struggling to prove the existence of the map $\varphi$. I suspect it might follow from some sort of universal mapping property, e.g. the UMP of the cokernel, but I can't find one which is relevant to this problem. Or perhaps I can say something like a smooth manifold is isomorphic to its image under an embedding, giving $\varphi$ as the composite: $$ H_{1}\xrightarrow{\sim} i_1(H_1) = i_2(H_2) \xrightarrow{\sim} H_2$ $$

Any guidance would be greatly appreciated.

Ri-Li
  • 9,466
  • 4
  • 56
  • 104
An Coileanach
  • 227

1 Answers1

1

Define a map as follows. For each $g \in i_{1}(H_{1}) = i_{2}(H_{2})$, by injectivity there are unique elements $g_{1} \in H_{1}$ and $g_{2} \in H_{2}$ which map to $g$. Define$\varphi (g_{1}) = g_{2}$. One may verify this map is a homomorphism, and the inverse homomorphism is defined similarly.

Nick L
  • 4,520
  • 1
    You still need to prove that $\varphi$ is smooth (so that it is an isomorphism of Lie groups, not just abstract groups). This requires a bit of work, see my answer here. – Moishe Kohan Mar 05 '18 at 17:55
  • Thank you both for your help. As a map of abstract groups this works very nicely. @MoisheCohen, with regard to your linked answer (which was very illuminating), the proof of smoothness seems to come from the fact that the projections from the fiber product of the immersed subgroups is a Lie group morphism, so I just want to clarify: is the map $\varphi$ as defined by Nick L, the same as the composition $p_2 \circ F^{-1} \circ i_1$ in the language of your linked answer (and so a composition of morphisms)? – An Coileanach Mar 05 '18 at 19:38
  • 1
    @AnCoileanach: Yes, of course. – Moishe Kohan Mar 05 '18 at 23:01