Munkres topology Exercise $2.22$ Question $1$

The latter picture is what I currently have for this one, But I have no faith that it is correct, I am having a hard time with composing functions. Need an answer to learn from.

Solution 1:

In knowing $H$ is a topological group, you're given that $$h: H \times H \to H; h(x,y)=xy$$

is continuous and also that $$g: H \to H; g(x)=x^{-1}$$ is continuous.

Then you have to show that the combo map

$$f: H \times H \to H; f(x,y)=xy^{-1}$$ is continuous too.

The obvious way is indeed to use compositions. To compute $f$ you first have to apply $g$ to the second component of the pair, leaving the first component alone and to the resulting pair we apply the multiplication $h$, so $$f = h \circ ( \pi_1 \times (g \circ \pi_2))$$ which is continuous as a composition of continuous maps. The right hand product map (which is $H \times H \to H \times H$) is continuous by the universal property of $H \times H$, as the "components" (i.e. compositions with the projections) are precisely $\pi_1$ and $g \circ \pi_2$, both continuous.

So $f$ is continuous.

If $f$ is continuous we just note that $g(x) = f(1,x)$ and use that to show $g$ is continuous from ( note $c_1: H \to H$ is the constant map with value $1$): $$g = f \circ (c_1 \times \pi_2)$$ and then, from $$h(x,y)= xy= x(y^{-1})^{-1} = f(x,y^{-1})=f(x,g(y))$$ we can write as before

$$h = f \circ (\pi_1 \times (g \circ \pi_2))$$ and see that also $h$ is continuous, given that we already saw $g$ was.