Image of Homomorphism of Lie groups This is exercise from Lee: Introduction to smooth manifolds.
Suppose $f \colon G \to H$ is homomorphism of Lie groups (real, finite-dimensional).
Q: Is image $Im(f) \subseteq H$ a Lie subgroup of H?
That is, is there topology and smooth structure on $Im(f)$ such that inclusion $Im(f) \hookrightarrow H$ is immersion, and such that induced operation $Im(f) \times Im(f) \hookrightarrow H \times H \to Im(f) \subseteq H$ is smooth.
The author of these notes says (last paragraph on the first page) that the answer is no, provides counterexample (dense line on torus), but the proof is ommited. Is it correct?
I need positive proof of that, to understand the following characterisation: Lie group admits faithfull finite-dimensional representation if and only if it is (isomorphic to) Lie subgroup of $GL(n,\mathbb R)$.
One more thing. It is easy to see that $Ker(f)$ is Lie subgroup of $G$ using nontrivial Closed subgroup theorem. Is there more direct proof?
Any help is very appreciated.
 A: Let $f\colon G\to H$ be a homomorphism of Lie groups as in the question. Via the natural group isomorphism, we can consider $\DeclareMathOperator{\im}{im}\im(f)$ as $G/\ker(f)$. Since $\ker(f)\subset G$ is a closed subgroup, $G/\ker(f)$ has a canonical Lie group structure. Using this to impose a Lie group structure on $\im(f)$, it becomes a Lie subgroup of $H$.
We may clarify this, by showing that the given definition of Lie subgroups coincides with the following: A subset $I\subset H$ of a Lie group $H$ is an (immersed) Lie subgroup, if there exists a Lie group $I'$ and an injective homomorphism of Lie groups $I'\to H$, such that $\im(I'\to H) = I$.
Suppose this for the moment. Then the induced Lie group homomorphism $\bar{f}\colon I' := G/\ker(f)\hookrightarrow H$ has the same image as $f$ and is injective by construction. Thus $\im(f)$ is an immersed Lie subgroup of $H$.
To show the equivalence of the definitions, everything is tautological, except the fact that injective Lie group homomorphisms are automatically immersions. Actually, I didn't knew the way you are describing to see this in your comment below your question; I'm used to reduce the question to the unit and then argue via the exponential map, which is a local diffeomorphism. Indeed, it suffices to show that the derivative of $\iota\colon I'\to H$ is injective at the unit $e$. But the exponential maps $\exp_{I'}\colon T_e I'\to I'$ and $\exp_{H}\colon T_e H\to H$ are local diffeomorphisms at $0$, and since we have a commutative square $\exp_{H}\circ d\iota = \iota\circ\exp_{I'}$, the injectivity of $d\iota$ at $e$ follows from the injectivity of $\iota$.
