This is a seemingly simple exercise in Chapter 3 of Lee's "Topological Manifolds", but it's making me feel like I didn't understand the material.
For completeness, here is the Corollary --- as written by Lee --- about "passing to the quotient":
Corollary 3.30 (Passing to the Quotient). Suppose $\pi:X \rightarrow Y$ is a quotient map, $B$ is a topological space, and $f: X \rightarrow B$ is any continuous map that is constant on the fibers of $\pi$ (i.e., $\pi\left(p\right) = \pi\left(q\right)$, then $f\left(p\right)=f\left(q\right)$). Then there exists a unique continuous map $\tilde{f}: Y\rightarrow B$ such that $f = \tilde{f}\circ \pi$.
This is basically saying that if we have a space $X$ and a quotient map $\pi$ from $X$ to a quotient space $Y$, then any continuous function on $X$ which is constant on the fibers of $\pi$ can sensibly be extended to be applied to the quotient space $Y$, and is still continuous.
Here is the problem:
3-14. Let $G$ be a topological group and $\Gamma \subset G$ be a subgroup. (a) For any $g \in G$, show that left translation $L_g: G \rightarrow G$ passes to the quotient $G/\Gamma$ and defines a homeomorphism of $G/\Gamma$ with itself.
When dealing with topological groups, a subgroup $\Gamma \subset G$ naturally induces a quotient map $$ \pi: G \rightarrow G/\Gamma \text{ defined by } \pi\left(g\right) = g\Gamma $$ sending each group element to its coset of $G$ by $\Gamma$.
For fixed $g\in G$, the left translation map $L_g: G \rightarrow G$ defined by $L_g\left(h\right) = gh$ is known to be continuous (actually a homeomorphism). But $L_g$ is certainly not constant on the fibers of $\pi$: let $\pi\left(x\right) = \pi\left(y\right)$, then $x\Gamma = y\Gamma$ and so $x = y\gamma$ for some $\gamma \in \Gamma$. But then $$L_g\left(x\right) = gx = gy\gamma = L_g\left(y\right)\gamma \neq L_G\left(y\right) $$ in general. So I don't understand how the left translation map "passes to the quotient" here.
Since the question mentions "defining a homeomorphism of $G/\Gamma$ with itself", I thought that maybe it meant that $L_g$ somehow induced a continuous map $G \rightarrow G/\Gamma$, then by the above Corollarly, this map would pass to the quotient and become a continuous mapping $G/\Gamma \rightarrow G/\Gamma$ (which I could then show to be a homeomorphism). But if that's the case, then why is the language confusing me here?
Indeed, note that you want to get a map $G/\Gamma\to G/\Gamma$, not $G/\Gamma\to G$. This is done naturally. Define $f:G\to G/\Gamma$ as the composition $\pi\circ L_g$. This map is constant on the fibers, because $x\Gamma=y\Gamma$ clearly implies $gx\Gamma=gy\Gamma$. So now $f$ induces a map from $G/\Gamma$ to itself. It is easy to show this is a homeomorphism.