I am struggling to prove the following straight-line homotopy continuous.
For continuous functions $f, g: I \rightarrow \mathbb{R}^2$, defined by $f(x) = \left(\cos(\pi x), -\sin(\pi x)\right)$ and $g(x) = \left(\cos(\pi x), \sin(\pi x)\right)$, there exists a homotopy $F:I \times I \rightarrow \mathbb{R}^2$, defined by $F(x, t) = \left(\cos(\pi x), (2t - 1) \, \sin(\pi x)\right)$.
Although I understand that $F(x, t)$ should be continuous since it is made up of compositions and products of continuous functions, I tried to use the general structure presented in this
answer to prove it rigorously, but finding the structure hard to understand, I couldn't make much progress.
Let's do the same thing:
Note that by the universal property of the product it suffices to show that each coordinate is continuous.
The first coordinate is the projection $I\times I \to I$ followed by $x \mapsto \cos(\pi x)$ which are both continuous, so it's continuous.
The second coordinate is $m \circ (h\times k)$ where $m :\mathbb R\times \mathbb R \to \mathbb R$ is the multiplication (which is continuous), $h: I\to \mathbb R$ is $t\mapsto 2t-1$, which is clearly continuous, and $k:I\to \mathbb R$ is $x\mapsto \sin(\pi x)$, which is also continuous. Hence $h\times k$ is continuous, and thus $m\circ (h\times k)$ is as well : the second coordinate is also continuous.