Define $H : I \times I \to \mathbb{C} \setminus \{0\}$ by $H(s, t) = e^{2\pi ist}$, show that $H$ is continuous

Question

Define $H : I \times I \to \mathbb{C} \setminus \{0\}$ by $H(s, t) = e^{2\pi ist}$, show that $H$ is continuous

The biggest problem for me in veryifying homotopies is proving that the claimed homotopy is actually continuous. This $H$ is a homtopy from the constant path $c(s) = 1$ to $f(s) = e^{2\pi is}$, I am trying to verify that it is continuous.

Intuitively I can see that it should be continuous, but verifying that it is continuous rigorously is proving to be a bit hard. I highly doubt that it would be easy to prove that it's continuous directly from the definition of continuity of a function between two topological spaces.

I tried the following. Let $\mathcal{B}'$ be a basis for $\mathbb{C} \setminus \{0\}$, then $\mathcal{B} = \{ B \cap \mathbb{C} \setminus \{0\} \ | \ B \in \mathcal{B}\}$, where $\mathcal{B}$ is the basis for the metric topology on $\mathbb{C}$, pick a basis element $U = B_{(\mathbb{C}, d)}(x, \epsilon) \cap \mathbb{C} \setminus \{0\}$ for some $x \in \mathbb{C}$ and some $\epsilon > 0$, but the problem is $H^{-1}[U]$ is undefined since there exists $y \in U$ for which there doesn't exists $x \in I \times I$ such that $H(x) = y$. This is because $H[I \times I] = \mathbb{S}^1 \subseteq \mathbb{C} \setminus \{0\}$. So I'm not sure how to verify continuity using the defintition of continuity of a function (along with the theorem that we only need to she that the inverse image of every basis element in the target space is open)

Is there an easier way to prove that $H$ is continuous? Furthermore is there a way I can prove $H$ is continuous using the definition of a continuous function between topological spaces? Finally, how does one normally go about verifying that Homotopies are continuous, are there a standard set of therems, (e.g. the gluing lemma) that one normally uses?

Answer 1

If I were teaching a course in which this was an exercise, I'd expect students to say things like

1. The function $z \mapsto e^z$ is analytic, hence continuous.
2. The function $(s, t) \mapsto st$ is a polynomial, hence differentiable, hence continuous in each variable.
3. The function $u \mapsto cu$ is continuous for every constant $c$, by a similar argument.

Hence the composite $(s, t) \mapsto e^{2\pi i st}$ is continuous. Indeed, I might just expect them to say "From calc/complex variables, we know that $(s, t) \mapsto ...$ is continuous because it's a composition of continuous functions."

Answer 2

With $e^{ix}=\cos x+i \sin x$ for real $x$ and

$|\cos x- \cos y| \le |x-y|$ and $|\sin x- \sin y| \le |x-y|$ for real $x,y$ (mean value theorem !) we get

$|H(s,t)-H(u,v)| \le 4 \pi |st-uv|$.