TITLE: Continuity of homotopy from constant function to identity
QUESTION [5 upvotes]: 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

REPLY [3 votes]: If $X$ is a subset of $\mathbb R^n$, you can use the fact that $\mathbb R^n$ is a topological group (actually a Lie group), so that the sum and product (product in $\mathbb R^n ; n>2$ means scalar product  $\mathbb R \times \mathbb R^n \rightarrow \mathbb R^n$) are both continuous operations.
NOTE: An important temptation to avoid is that of trying to show the continuity of $f(x,y)$ by showing the continuity of both $f(x,.)$ and $f(.,y)$ separately, since this does not imply the continuity of $f(x,y)$.