Prob. 5 (a) in Supplementary Exercises in Munkres' TOPOLOGY, 2nd ed: How to show that this map is a homeomorphism?

Question

Let $G$ be a topplogical group, and let $H$ be a subgroup of $G$. Let $G / H $ denote the collection of all left cosets of $H$ in $G$, and let $a$ be a fixed element of $G$. Let the map $f \colon G / H \to G / H$ be defined as follows: $$f(xH) \colon= axH \ \ \ \mbox{ for all } \ xH \in G/H.$$

Then how to show that the map $f$ is a homeomorphism?

Here $G/H$ has the quotient topology.

My effort:

For any two elements $x, y \in G$, we have $xH = yH$ if and only if $x^{-1}y \in H$.

So if $xH = yH$, then $x^{-1}y \in H$; so $$(ax)^{-1}(ay) = ( x^{-1}a^{-1})(ay) = x^{-1}y \in H,$$ showing that $axH = ayH$. So $f$ is well-defined (i.e. single-valued).

Now if $f(x) = f(y)$, then $axH = ayH$ and so $(ax)^{-1}(ay) \in H$. But $$(ax)^{-1}(ay) = ( x^{-1}a^{-1})(ay) = x^{-1}y. $$ So $x^{-1}y \in H$, which implies that $xH = yH$. Thus $f$ is injective.

Now if $yH \in G/H$, then $y \in G$ and so $a^{-1}y \in G$ also; therefore $a^{-1}y H \in G/H$. Moreover, $$f(a^{-1}yH) = aa^{-1}yH = yH,$$ showing that $f$ is surjective.

Hence $f$ is bijective.

How to proceed from here? How to show that $f$ is continuous? And, how to show that $f$ is open?

I know that the map $f_a \colon G \to G$ defined by $$f_a(x) \colon= ax \ \ \ \mbox{ for all } \ x \in G$$ is a homeomorphism.

And, the map $p \colon G \to G/H$ defined by $$p(x) \colon= xH \ \ \ \mbox{ for all } \ x \in G$$ is surjective.

Moreover, the action of the map $f$ on the elements of $G/H$ is in some way similar to the action of the map $f_a$ on the elements of $G$.

How to proceed from here?

Answer 1

Let us first show that $f$ is continuous: given $U \subset G/H$ open, $f^{-1}(U)$ is open iff $p^{-1}(f^{-1}(U))$ is open, where $p \colon G \to G/H$ is the quotient map. However, $$ p^{-1}(f^{-1}(U)) = (f \circ p)^{-1}(U) = (p \circ \tilde{f})^{-1}(U) $$ is open, where $\tilde{f} \colon G \to G$ is left-multiplication by $a$. This is because $p \circ \tilde{f} \colon G \to G \to G/H$ is a composition of continuous maps, hence continuous. Moreover, the same argument shows that the map $g \colon G/H \to G/H$ given by $xH \mapsto a^{-1} xH$ is continuous, and it is clearly the inverse of $f$, so $f$ is a homeomorphism.

Answer 2

To finish the argument and show that we have a homeomorphism $G/H \mapsto G/H$, you use corollary 22.3 in Munkres, chapter 2 section 22. To summarize and translate the relevant parts of the corollary to this exercise:

Let $p$ be the standard quotient map $G \mapsto G/H$, and $f_{\alpha}: G \mapsto G$ be defined as in Munkres, exercise 4 (= $x \mapsto \alpha x$). We have a quotient map $g: X \mapsto Z$ (this is the map $$g :=(p \circ f_{\alpha}): G \mapsto G/H$$ above), and a set $X^* := \{g^{-1}({z}), z \in Z\}$ endowed with the quotient topology. The latter translates to $$X^* = \{(p \circ f_{\alpha})^{-1}(xH), xH \in G/H\} = G$$,

which we endow with the quotient topology, so it's $G/H$ again.

Then if $g$ is a quotient map (check), there is a homeomorphism $$h:G/H \mapsto G/H$$ (namely, $h = g\circ p^{-1} = p \circ f_{\alpha} \circ p^{-1}$).