Can't see why particular homotopy is continuous

Question

I'm checking the group laws for the fundamental group of $(X,x_{0})$: in particular I'm trying to show that $\gamma \simeq \gamma \cdot e$ , where $\gamma$ is a loop based at $x_{0}$, $e$ is the constant path and $\cdot$ denotes concatenation. The obvious homotopy to write down is: $$\begin{array}{lclr} H(t,s)&=&\gamma((1+s)t)&t \in[0,\frac{1}{1+s}] \\ &&x_{0}&t \in [\frac{1}{1+s},1] \end{array}$$ Then $H(t,0)=\gamma(t)$ and $H(t,1)=\gamma \cdot e$.

I'm having trouble convincing myself that $H$ is continuous. It's clear that $H(\cdot,s)$ is continuous for each $s$, but I'm not sure how to show that $H$ itself is continuous.

Answer 1

Let $U \subset X$ be an open set. Then $\gamma^{-1}(U) = V$ is open since $\gamma$ is continuous.

We have that $H^{-1}(U)$ is the union of $\{ (t, s) \in I \times I: \gamma((1 + s) t) \in U \}$ and $\{(t, s) \in I \times I: t > \frac{1}{1 + s} \}$ if $x_0 \in U$. Note that we can take strict inequality here because of the way that $H$ is patched together.

The first set is equal to $\{ (t, s) \in I \times I: (1 + s) t \in V \}$, which is open since $(t, s) \mapsto (1 + s) t$ is continuous and $V$ is open.

Similarly, the second set is open becuase it is the pre-image of $(-\infty, 0)$ w.r.t. the continuous function $(t, s) \mapsto \frac{1}{1 + s} - t$, $(t, s) \in I \times I$.

Thus $H^{-1}(U)$ is open and hence $H: I \times I \rightarrow X$ is continuous.