Let $G$ be a topological group. Let $G_0 = \{g \in G\ $ $: e$ and $g$ are in the same path connected component of $G$$\}.$ Show that $G_0$ is a normal subgroup of $G.$
My attempt $:$ First I will prove that the left multiplication by any element of $G$ is continuous. Let $a \in G$ be a fixed element. First we consider the map $\varphi : G \longrightarrow G \times G$ defined by $g \longmapsto (a,g),\ g \in G.$ Then $\varphi$ is continuous on $G.$ To see that we take an open subset $U = \bigcup\limits_{i \in I} \left (U_i \times V_i \right )$ in $(G \times G),$ where $U_i,V_i$ are open subsets of $G$ for each $i \in I$ (some index set). Then $\varphi^{-1} (U) = \bigcup\limits_{\substack {i \in I \\ a \in U_i}} V_i,$ which is clearly open in $G$ for being the union of arbitrary open subsets of $G.$ Hence $\varphi$ is continuous. Since $G$ is a topological group the map $\psi : G \times G \longrightarrow G$ defined by $(g,g') \mapsto gg',\ (g,g') \in G \times G$ is also continuous. So the map $(\psi \circ \varphi) : G \longrightarrow G$ defined by $(\psi \circ \varphi) (g) = ag,\ g \in G$ is also continuous for being the composition of two continuous maps. Thus left multiplication by any element of $G$ is continuous (in fact a homeomorphism on $G$), as claimed. Similarly we can prove that the right multiplication by any element of $G$ is also continuous. From now on I will use $L_a$ to denote the left multiplication by $a \in G$ and $R_b$ to denote the right multiplication by $b \in G.$
Now let us prove that $G_0$ is a subgroup of $G.$ Let us take $g,g' \in G_0.$ Then there exist paths $\gamma_1$ joining $e$ and $g$ and $\gamma_2$ joining $e$ and $g'.$ Then $(R_{g'} \circ \gamma_1)$ defines a path from $g'$ to $gg'.$ Therefore the concatenation $\gamma_2 + (R_{g'} \circ \gamma_1)$ defines a path from $e$ to $gg'.$ This shows that $gg' \in G_0.$ Also since $G$ is a topological group the map $\iota : G \longrightarrow G$ defined by $g \longmapsto g^{-1}$ is continuous (in fact a homeomorphism on $G$). Now for any $g \in G_0$ let $\gamma$ be a path joining $e$ and $g.$ Then $(\iota \circ \gamma)$ defines a path from $e$ to $g^{-1},$ proving that $g^{-1} \in G_0.$ This shows that $G_0$ is a subgroup of $G.$
To prove that $G_0$ is a normal subgroup of $G$ we take $g \in G$ and $g_0 \in G_0.$ Need only to show that $gg_0g^{-1} \in G_0.$ Since $g_0 \in G$ there exists a path $\gamma_0$ joining $e$ and $g_0.$ Then $(R_{g^{-1}} \circ L_g \circ \gamma_0)$ defines a path from $e$ to $gg_0g^{-1},$ proving that $gg_0g^{-1} \in G_0,$ as required.
This completes the proof.
QED
Can anybody please check my solution? Thanks for your time.