Is an injective morphism from a Lie group to itself surjective? I have a question about Lie groups. Let $G$ be a finite dimensional (real or complex) Lie group and $f:G \rightarrow G$ an homomorphism of Lie groups. Edit : in the view of some counter-examples let's also suppose $G$ connected.
Q1 : Suppose that $f$ is injective, is it then surjective ?
Here are my attempts so far to adress this question. It is known that an injective Lie homomorphism is an immersion so the differential $df_g$ of $f$ in any point $g\in G$ is injective. For reasons of dimension, $df_g$ is then bijective. Thus, by the inversion theorem, $f$ is a diffeormorphism from $G$ to $f(G)\leq G$.
Moreover, it is also well known that $f(G) \simeq G/Ker(f) \simeq G/\lbrace e \rbrace \simeq G$. So we have a subgroup $f(G)$ of $G$ isomorphic to $G$. Then, we can reduce the question to :
Q2 : Let $H$ be a Lie subgroup of $G$. If $H \simeq G$, do we have $H=G$ ?
Now, maybe we can progress with some topological argument. Typically, I think of the fundamental group, that is useful to link the classification of Lie groups to the classification of Lie algebras. As $H$ is diffeomorphic to $G$, we have $\pi(H)=\pi(G)$. Hence a third question :
Q3 : Let $H$ be a Lie subgroup of $G$. If $Lie(H)=Lie(G)$ and $\pi(H)=\pi(G)$, do we have $H=G$ ?
After that, I have to admit that I don't have ideas about where to look. So if someone as an answer, a hint or a reference, that would be awesome.
In addition, I have a few "bonus" questions around this subject :
B1 : If the answer to the questions above is no, what is a counterexample ? Is there a simple hypothesis we can add so that it becomes true ?
B2 : What are the possible generalisations to all this discussion ? For example with infinite dimensional Lie groups ? With topological groups ?
B3 : In the other way around, if $f\in Hom(G,G)$ is surjective, is it injective ?
Thanks a lot to all who have read till the end and to the ones who will answer.
 A: For the sake of having an answer, and to summarize some discussion in both the question and the comments:
If $G$ is connected and $f : G \to G$ is injective, then it is an immersion, hence a local diffeomorphism, hence open. But the only open subgroup of a connected topological group $G$ is $G$ itself, since any open neighborhood of the identity in $G$ generates $G$. Hence $f$ is surjective. 
So with connectivity, the answers to Q1, Q2, Q3 are all yes, and in Q3 the extra hypothesis on fundamental groups is unnecessary. 
The answer to a slightly different version of Q3 is no: you might ask whether two connected Lie groups are isomorphic if their Lie algebras and fundamental groups are isomorphic. As a counterexample, $SL_2(\mathbb{R})$ has fundamental group $\mathbb{Z}$, and hence it admits a unique connected double cover whose fundamental group is also $\mathbb{Z}$ (and which also has Lie algebra $\mathfrak{sl}_2(\mathbb{R})$), the metaplectic group $Mp_2(\mathbb{R})$. These two Lie groups are not isomorphic: the metaplectic group has no faithful finite-dimensional representations. 
It's a bit trickier to find a compact counterexample, but it can still be done: $SU(2) \times SU(2)$ double covers three Lie groups, namely $SU(2) \times SO(3), SO(3) \times SU(2)$, and $SO(4)$. All of these have the same Lie algebra (namely $\mathfrak{su}(2) \times \mathfrak{su}(2)$) and fundamental group (namely $\mathbb{Z}_2$) but $SO(4)$ is not isomorphic to $SU(2) \times SO(3)$: their representation theory is different. 
The answer to another slightly different version of Q3 is yes: you might ask whether a map $f : G \to H$ of connected Lie groups is an isomorphism if it induces an isomorphism on Lie algebras and on $\pi_1$. This is true and follows from covering space theory, since the first hypothesis implies that $f$ induces an isomorphism on universal covers. 
