Exact sequence of topological groups induces exact sequence on the fundamental group I think it’s a general result, but if you like you can assume compact Lie Groups.
Here is the situation:
We have a exact sequence of path connected topological groups
$$ 1 \rightarrow  A \rightarrow B \rightarrow C \rightarrow 1$$
Name the continuous homomorphisms $f: A \rightarrow B$ and $g: B \rightarrow C$. I would like to know if the following sequence is exact:
$$ 1 \rightarrow \pi_1(A) \rightarrow \pi_1(B) \rightarrow \pi_1(C) \rightarrow 1$$
With the induced maps $f_*$ and $g_*$. 
I already proved $\operatorname{Im} f_* \subset \ker g_*$, but I can’t conclude the other inclusion. If you have any other suggestion to conclude it I would be very happy.
Remark: In general it’s not true that a injective map implies that the induced map on the fundamental group is injective. Example: inclusion of $\mathbb{S}^1$ in $\mathbb{B}^1$.
 A: I am supposing $A$, $B$ and $C$ as compact connected Lie Groups.
Since
$$1\longrightarrow A \xrightarrow{\ \ f \ \ }B \xrightarrow{\ g \ }C\longrightarrow 1,$$
is exact, $f$ is an injection. Using that $A$ is a compact Lie group one can conclude that $f(A)$ is a closed subgroup of $B$, and therefore $f$ is an embedding by Cartan's theorem which shows us that $f(A)$ is an embedded submanifold of $B$. Note that $f(A)=\text{Ker}(g)$ $\Rightarrow$ $f(A)$ is a normal subgroup of $B$. Moreover, $C\cong B/\text{Ker(g)} = B/\text{Im}(f),$ by the map $\tilde{g}: B/\text{Im}(f) \to C$, $\tilde g(b+\text{Im}(f)) = g(b)$. Since $f(A)$ is closed then   $B/\text{Im}(f)$ is also a Lie group.
And then, we have the comutative diagram
$$\require{AMScd}
\begin{CD}
1 @>>>A @>f>> B @>g>> C@>>> 1\\
  @.      @VV{f}V @VV{\text{Id}}V  @VV{\tilde{g}^{-1}}V @.\\
1@>>> f(A) @>\iota>> B @>\pi>> B/f(A)@>>>1
\end{CD}.$$
It is clear that the following induced diagram is also commutative
$$\require{AMScd}
\begin{CD}
1 @>>>\pi_1(A) @>f_*>> \pi_1(B) @>g_*>> \pi_1(C)@>>> 1\\
  @.      @VV{f_*}V @VV{\text{Id}_*}V  @VV{\tilde{g}^{-1}_*}V @.\\
1@>>> \pi_1(f(A)) @>\iota_*>> \pi_1(B) @>\pi_*>> \pi_1(B/f(A))@>>>1
\end{CD} ,$$
Thus it is enough to prove that the last line of the above diagram is an exact sequence ($f_*$, $\text{Id}_*$ and $\tilde{g}^{-1}_*$ are isomorphisms).
But this is obvious since $f(A) \to B \to B/f(A)$ is a fibration and from the long exact sequence of fibration
$$\ldots\to \pi_n(f(A))\to \pi_n(B)\to \pi_n(B/\text{Im}(A))\to\ldots\to \pi_2(B/\text{Im}(A))\to \pi_1(f(A))\to \pi_1(B)\to \pi_1(B/\text{Im}(A))\to {1}, $$
since $\pi_2(B/\text{Im}(f)) =1$, because $B/\text{Im}(f)$ is a Lie group we have proved the result.
Remark: We have supposed $A$, $B$ and $C$ as Lie groups just to be able to conclude that $\pi_2(B/\text{Im}(f))$ is trivial.
EDIT: As Aloizio Macedo pointed out in the comment section, it is also important to suppose $A$, $B$ and $C$ Lie groups in order to guarantee that we have a fibration.
