Every continuous function $\mathbb{R}^n\rightarrow X$ is null-homotopic

Question

Let $f:\mathbb{R}^n\rightarrow X$ be continuous and $g:\mathbb{R}^n\rightarrow X$ a constant map with image $\{x_0\}$.
For the proof, I thought that we can define $F:[0,1]\times\mathbb{R}^n$ with $F(s,x)=(1-s)f(x)+sx_0$. Is this the homotopy we're looking for? It seems rather easy, or am I missing something?

Answer 1

You need $x_0$ to be in the same path component as $Im f$

$F(t,x) := f(tx)$ gives you a homotopy between $f$ and the constant map at $f(0) \in X$. Now take a path $h: [0,1] \to X$ from $f(0)$ to $x_0$. Define a functtion $H(t,x) : = h(t)$. This gives you the homotopy between the constant map at $f(0)$ and $g$. So you are done by transitivity of homotopy.