Unclear step in universal complexification proof (isomorphism of fundamental group induced by inclusion) I am stuck with the proof of the following theorem from Lie groups by Daniel Bump (2nd edition).

Theorem 24.1. Let $K$ be a compact connected Lie group. Then $K$ has a complexification $K \to G$, where $G$ is a complex analytic group. The induced map $\pi_{1}(K) \to \pi_{1}(G)$ is an isomorphism. The Lie algebra of $G$ is the complexification of the Lie algebra of $K$. Any faithful complex representation of $K$ can be extended to a faithful analytic complex representation of $G$. Any analytic representation of $G$ is completely reducible.

There is a step in the proof in the book that I can't understand (I am not reporting the full demonstration here for the reader's convenience but the next paragraph is a summary):
$G$  is taken as $PK$, where $P=\{e^{iX}|X \in \kappa\}$  and $\kappa$  is the Lie algebra of $K$. $K$ is assumed to be unitary, with no loss of generality.
Then $G=PK$ is proven to be a closed subgroup of $GL(n, \mathbb{C})$.
I am now quoting the remainder of the demostration below, highlighting in italics the step that I don't understand.

Since $P$ is homeomorphic to a vector space, it is contractible, and since $G$ is homeomorphic to $P \times K$, it follows that the inclusion $K \to G$ induces an isomorphism of fundamental groups.
The Lie algebra of $G$ is, by construction, $i\kappa + \kappa = \kappa_{\mathbb{C}}$.
To show that $G$ is the complexification of $K$, let $H$ be a complex analytic group and $f : K \to H$ be a Lie group homomorphism. We have an induced homomorphism $\kappa \to Lie(H)$ of Lie algebras, which induces a homomorphism $k_{\mathbb{C}}= Lie(G) \to Lie(H)$ of complex Lie algebras, by [...]. If $\tilde{G}$ is the universal covering group of $G$ then by Proposition 14.2* we obtain a Lie group homomorphism $\tilde{G} \to H$. To show that it factors through $G \cong \tilde{G}/\pi_1(G)$, we must show that the composite $\pi_1(G) \to \tilde{G} \to H$ is trivial.
But this coincides with the composition $\pi_1(G)\cong \pi_1(K) \to \tilde{K} \to K \to H$  , where $\tilde{K}$ is the universal covering group of $K$, and the composition $\pi_1(K) \to \tilde{K} \to K \to H$ is already trivial. Hence the map $\tilde{G} \to H$ factors through $G$, proving that $G$ has the universal property of the complexification. [...]


*Proposition 14.2 Let $G$ and $H$ be Lie groups with Lie algebras $\mathfrak{g}$ and $\mathfrak{h}$, respectively, and let $\pi : \mathfrak{g} \to \mathfrak{h}$  be a Lie algebra homomorphism. Then there exists a neighborhood $U$ of $G$ and a local homomorphism $\pi : U \to H$ whose differential is $\pi$.

and also

Theorem 14.2. Let $G$ and $H$ be Lie groups with Lie algebras $ \mathfrak{g}$ and $\mathfrak{h}$, respectively, and let $\pi : \mathfrak{g} \to \mathfrak{h}$ be a Lie algebra homomorphism. Assume that $G$ is simply connected. Then there exists a Lie group homomorphism $\pi : G \to H$ with differential $\pi$.

I understand that we need to prove that $\pi_1(G)$ is in the kernel of $\tilde{G} \to H$ to apply the fundamental theorem of homomorphism and show that the homomorphism factors through $G=\tilde{G}/\pi_1(G)$, however I have the following questions:
Questions:

*

*I do not understand why the composition $\pi_1(G) \to \tilde{G} \to H$ coincides with $\pi_1(K) \to \tilde{K} \to {K} \to H$ i.e. if $a \in \pi_1(G)$ goes to $h_1$ why its corresponding $a \cong b \in \pi_1(K)$ goes to $h_1$ as well? I understand that $K \to G$ induces an isomorphism of the corresponding fundamental groups*, but I can't find a sequence of arrows, i.e. a commutative diagram, that proves the equivalence of compositions above. As I wrote in the comments, I think that, the homotopy class $c_k$ of loops of $K$ which is isomorphic to the homotopy class $c_g$ of $G$ is also included in $c_g$ but I can't see how to move from here (I just get $\pi_1(K) \to \tilde{G} \to G \to H$).


*Once we have ascertained that $K \to H$ induces $G \to H$, why can we say that $G \to H$ restricts to $K \to H$?


*In general, if $K \to G$ is an inclusion, what can be said about their universal coverings $\tilde{K}$ and $\tilde{G}$ ?
Any help is appreciated!
*My informal understanding of this induced isomorphism is that the inclusion $K \to G$ maps loops in $K$ to loops in $G$ and,as $P$ is contractible, any loop in $G=P \times K$ can be "projected" into to a loop in $K$. This should allow us to establish a 1 to 1 mapping between homotopy equivalent classes of loops in $G$ and homotopy equivalent classes of loops in $K$, aka between their fundamental groups. But in the demostration above, an element of the fundamental group is seen an element of the kernel of the projection $\pi: \tilde{G} \to G$ and I don't see how the induction by $K \to G$ works in this case.
 A: I will attempt a partial answer myself. Any suggestion or amendment is more than welcome.
As for question 1, I think we can say that, by definition, we have built two homomorphisms
$$\phi_1 : \tilde{K} \to K \to H$$
$$\phi_2 : \tilde{K} \to \tilde{G} \to H$$
that have the same differential $\mathfrak{\kappa} \to Lie(H) = \mathfrak{\kappa} \to \mathfrak{\kappa}_{C} \to Lie(H)$.
As $\tilde{K}$ is simply connected, the two homomorphisms coincide.
The inclusion $K \to G$ also induces the isomorphism between $\pi_1(K)$ and $\pi_1(G)$ which is the restriction of the induced homomorphism $\tilde{K} \to \tilde{G}$ to $\pi_1(K) \in \tilde{K}$.
From this, it follows that $\tilde{G} \to H$ brings all elements of $\pi_1(G)$ to $e$.
For question 2, similarly I think that taking an element $k \in K$ we need to prove that the homomorphism $K \to H$ and the induced homomorphism $G \to H$ bring $k$ to the same $h \in H$. If we take $p^{-1}(k) \in \tilde{K}$ as the fiber over $k$ of the covering map, we also have     that $\phi_2(p^{-1}(k))=h$. Again, $\tilde{K} \to \tilde{G}$ and $\tilde{G} \to H$ factor through $K$ and $G$ respectively, hence the image of $k$ under the inclusion $K \to G$ followed by $G \to H$ is again $h$.
