On Prop 1.3.2 in the context "Symmetry, representations, and invariants", I cannot understand the last sentence. Why does it explain that $G^\circ$ is normal? Can you explain me in more detail.
Prop. Let $G$ be a topological group. Then the identity componet of G is a normal subroup.
Proof. Let $G^\circ$ be the identity component if $G$. If $h \in G^\circ$, then $h \in L_h(G^\circ)$ because $e \in G^\circ$. (Here, $L_h : G \rightarrow G$ is the translation map defined by $L_h(x) = hx$. It is a homeomorphism). Since $L_h$ is a homeomorphism and $G^\circ \cap L_h(G ^\circ) \neq \emptyset$, $L_h(G^\circ) = G^\circ$, showing that $G^\circ$ is closed under multiplication. Since $e \in L_h(G^\circ)$, $h^{-1} \in G^\circ$, and so $G^\circ$ is a subgroup. If $g \in G$, the inner automorphism $\tau(g)$ is a homeomorphism that fixes $e$ and hence maps $G^\circ$ into $G^\circ$.
You can use the same argument as you used with multiplication.
Consider the homeomorphism $\tau(g)$ of $G$. Since $\tau(g)$ fixes $e$, we have $\tau(g)(G^{\circ}) \cap G^{\circ} \neq \emptyset$, and so $\tau(g)(G^{\circ}) \subseteq G^{\circ}$. Since this holds for all $g \in G$, the subgroup $G^{\circ}$ is normal in $G$.