Let $X$ be a topological space and let's denote $\pi_1(X,1)$ for it's Fundamental Group of loops at base 1. I'm going through the proof that Fundamental Group is indeed a group and I have a question about the proof of the inverse axiom from Hatcher's "Algebraic Topology".
Let's denote $I = [0,1]$ and fix a path $f:I \rightarrow X$ (Hatcher shows this on more generally on paths instead of loops). Let's define $\overline{f}:I \rightarrow X$ as $$\overline{f}(s) := f(1-s).$$ We want to show that we have the following equality for homotopy classes: $[f][\overline{f}]=[e_0]$, where $e_0$ indicates the constant path at $f(0)$. The idea is the following: we define $H(t,s): I \times I \rightarrow X$ by $$H(t,s) = \begin{cases} f(2s), &0\leq s \leq \frac{1}{2} - \frac{t}{2}\\ f(1-t), &\frac{1}{2}-\frac{t}{2} < s \leq \frac{1}{2}+\frac{t}{2}\\ f(2-2s), &\frac{1}{2}+\frac{t}{2}<s \leq 1. \end{cases}$$ This $H(t,s)$ is the key since this will be a homotopy between $f \cdot \overline{f}$ (where $\cdot$ indicated product of paths) and $e_0$, since $H(0,s) = f \cdot \overline{f}(s)$ and $H(1,s) = e_0$, giving us our result.
My question is about the continuity of this function, to actually show that it's a homotopy. It's easy to see from the definition that $H(t,s)$ is continuous for each $t$ if we fix $s$, and for each $s$ if we fix $t$, but if I understand correctly the whole point is for $H(t,s)$ to be continous with respect to $(t,s)$ "at the same time". Honestly, I don't really feel what is the idea behind showing that $H(t,s)$ is continuous with respect to $(t,s)$, since the position of $s$ depends on $t$ in the definition, so I'm not exactly sure what kind of limits I should study (or perhaps this question is just silly and trivial?). I would really appreciate some insight here.
Thanks!