I'm following Lee's "Introduction to Topological Manifolds" and
3-16. If $\Gamma$ is a normal subgroup of the topological group $G$, show that the coset space $G/\Gamma$ is a topological group.
Below I provide my solution/proof of this statement. My confusion is that the author offers a hint pointing to previous exercises which show that (a) a finite product of open maps is open and (b) the quotient map induced by a continuous group action is an open map. Nowhere in my proof am I using either of these facts, so am I going wrong somewhere? I seemingly am proving the result using just the notion of "passing to the quotient."
My Proof
Let $\Gamma$ be a normal subgroup of topological group $G$. Then the usual inversion and group multiplication operations on $G$, $$ \iota: G\rightarrow G \text{ given by } \iota\left(g\right) = g^{-1} $$ and $$ \mu: G \times G \rightarrow G \text{ given by } \mu\left(\left(g,h\right)\right) = gh, $$ are continuous. We know from elementary abstract algebra that since $\Gamma$ is normal, the quotient group $G/\Gamma$ is itself a group under the operations $$ \overline{\iota}: G/\Gamma \rightarrow G/\Gamma \text{ given by } \overline{\iota}\left(g\Gamma\right) = g^{-1}\Gamma $$ and $$ \overline{\mu}: G/\Gamma \times G/\Gamma \rightarrow G/\Gamma \text{ given by } \overline{\mu}\left(\left(g\Gamma, h\Gamma\right)\right) = \left(gh\right)\Gamma. $$ To show that $G/\Gamma$ is a topological group, it remains to show that these operations are continuous with respect to the quotient topology on $G/\Gamma$. Lastly, let $\pi: G \rightarrow G/\Gamma$ be the standard projection map $\pi\left(g\right) = g\Gamma$ for $g\in G$.
First, the inversion map. Let $$\psi = \pi \circ \iota: G \rightarrow G/\Gamma \text{ be given by } \psi\left(g\right) = g^{-1}\Gamma $$ which is clearly continuous as the composition of two continuous maps. Then it's easy to see that $\psi$ is constant on the fibers of $\pi$:
$$ \pi\left(x\right) = \pi\left(y\right) \iff x\Gamma = y\Gamma
\iff x = \gamma y \text{ for some } \gamma \in \Gamma $$
(since $\Gamma$ is a normal subgroup!) hence $$ \psi\left(x\right) = x^{-1}\gamma = \left(\gamma y\right)^{-1}\Gamma = y^{-1} \gamma^{-1} \Gamma = y^{-1} \Gamma = \psi\left(y\right).$$ Thus, $\psi$ passes to the quotient to give rise to a (unique) continuous map $\overline{\psi}: G/\Gamma \rightarrow G/\Gamma$ such that $ \overline{\psi} \circ \pi = \pi \circ \iota = \psi$; it is obvious that $\overline{\psi} = \overline{\iota}$, the inversion map of the group $G/\Gamma$, so inversion is continuous.
Next, the multiplication map. The procedure is very similar, except the spaces are different, so I will use similar notation. Let $$ \psi = \pi \circ \mu: G \times G \rightarrow G/\Gamma $$ which is continuous as the composition of continuous maps. Define $\pi \times \pi: G\times G \rightarrow G/\Gamma \times G/\Gamma$ as the product of "two copies" of the quotient map on $G$; this map is a quotient map, being surjective and continuous (as the product of two continuous maps). Then $\psi$ is constant on the fibers of $\pi\times \pi$: $$ \pi \times \pi\left(\left(g,h\right)\right) = \pi\times\pi\left(\left(g',h'\right)\right) \implies g = \gamma g', h = h' \overline{\gamma} \text{ for some } \gamma, \overline{\gamma} \in \Gamma.$$ Then $$ \psi\left(\left(g,h\right)\right) = gh\Gamma = \gamma g'h' \overline{\gamma} \Gamma = g'h' \Gamma = \psi\left(\left(g',h'\right)\right) $$ where we used the fact that $\Gamma$ is normal to "absorb" $\gamma$ and $\overline{\gamma}$. Thus, $\psi$ passes to the quotient and induces a (unique) continuous map $\overline{\psi}: G/\Gamma \times G/\Gamma \rightarrow G/\Gamma $ such that $ \overline{\psi} \circ \left(\pi \times \pi\right) = \pi \circ \mu = \psi $. Then clearly $\overline{\psi} = \overline{\mu}$, the multiplication map of the group $G/\Gamma$, and so this multiplication is continuous.
Thus, $G/\Gamma$ is a topological group since both inversion and the group multiplication operation are continuous (with respect to the quotient topology).
Question Are there any major gaps in my proof? The hints about "open maps" is making me second-guess my logic.
Best Answer
I've come back to studying these concepts and finally understand what I was missing here. I will correct this proof for the multiplication operation.
First I will show that the map $p=\pi \times \pi: G \times G \rightarrow \left(G/\Gamma\right) \times \left(G/\Gamma\right)$ is a quotient map. It is obviously surjective and continuous, as the product of two surjective and continuous maps. (Here is what I missed previously:) The quotient map $\pi: G\rightarrow G/\Gamma$ is open (from Problem 3-15a) and so $p$ is open as the finite product of open maps (from Problem 1), and so $p$ is a quotient map.
The fibers of $p$ are of the form $$ p^{-1}\left(a\Gamma, b\Gamma\right) = a\Gamma \times b\Gamma. $$ Note that the mapping $\pi \circ \mu: G\times G \rightarrow G/\Gamma$ is constant on these fibers: if $\left(x,y\right), \left(u,v\right) \in p^{-1}\left(a\Gamma, b\Gamma\right) =a\Gamma \times b\Gamma$, then $x = \gamma u, y = v\gamma'$ for some $\gamma, \gamma'\in \Gamma$, and so $$ \pi\circ \mu\left(x,y\right) = xy\Gamma = \gamma uv \gamma' \Gamma = uv \Gamma = \pi \circ \mu\left(u,v\right) $$ since $\Gamma$ is a normal subgroup. Thus is is constant on these fibers. $\pi \circ \mu$ is continuous as the composition of continuous maps. Hence we can pass this to the quotient to obtain the continuous map $\overline{\psi}:\left(G/\Gamma\right) \times \left(G/\Gamma\right) \rightarrow G/\Gamma$ given by $\overline{\psi}\left(a\Gamma, b\Gamma\right) = ab\Gamma$ which is exactly the multiplication map on $\left(G/\Gamma\right)\times \left(G/\Gamma\right)$, hence the latter is continuous.
Thus, both operations are continuous and so $G/\Gamma$ is a topological group.