According to the definition of homotopy between two maps $f,g : X \to Y$ we need a continuous map $F : X \times [0,1] \to Y$ such that $F(x,0) = f(x)$ and $F(x,1) = g(x)$. Most examples I've seen in the literature however tend to skip over the continuity part of $F$ in examples.
So for example take $X$ a convex subset of $\mathbb{R}^n$, fix $x_0 \in X$ and define maps $X \to X$ by $f(x) = x_0$, i.e. $f$ is the constant function and the identity map $\mathrm{id}$. Then for a homotopy between $f$ and $\mathrm{id}$ we clearly want $F(x,t) = tx_0 + (1-t)x$ (obviously could generalise this to any $f$ and $g$). It seems non-trivial to me to prove that $F$ is continuous (the rest of it is straight forward).
I can formally solve to get an $F^{-1}$ for this but I'm not sure how that might help. I get $$ F^{-1}(y) = \left\{ \begin{array}{rl} \{ (x, 1) : x \in X \} \cup \{ (x_0, t) : t \in [0,1) \} & \text{if } y = x_0 \\ \left\{ \left( \frac{y-tx_0}{1-t}, t \right) : t \in [0,1) \right\} & \text{otherwise.} \end{array} \right. $$ but it seems tricky to then apply this to say the open ball about $x_0$ in $X$ say.
I also thought maybe trying to break it down into simpler maps but I can't see this either at the moment, for example I can't break it into maps $X \to X$ and $[0,1] \to X$ as then their product would be a map $X \times [0,1] \to X \times X$ which is of course not what I want.
Any help appreciated. I have a feeling it's not difficult I just can't see it right now.
Steve