How to build homotopy from convex to a point?

Question

How to build homotopy from convex to a point?

for example I want to build homotopy from the real line $\Bbb R $ to a single point $x_0$ I know that every path in a closed set $[a,x_0]$ in $\Bbb R$ is from the equation $a\cdot t+x_0\cdot (1-t) $ when $ t \in [0,1] $ but how to do this for all the real line?

so this will imply that the corresponde fundemental group in a convex is the trivial one.

Answer 1

Let $C$ be a convex subset of some $\mathbb{R}^n$ with the subspace topology inherited from $\mathbb{R}^n$. I guess you are looking for a homotopy $H : C \times [0,1] \to C$ such that $H(x,0) = x$ and $H(x,1) = x_0$ for all $x \in C$ (where $x_0 \in C$ is any prescribed point).

Define

$$H^\ast : \mathbb{R}^n \times [0,1]\to \mathbb{R}^n, H^\ast(x,t) = (1-t)x + tx_0 .$$

$H^\ast$ is easily seen to be a continuous map such that $H^\ast(x,0) = x$ and $H^\ast(x,1) = x_0$ for all $x$ . Moreover, we have $H^\ast(x,t) \in C$ for all $(x,t) \in C \times [0,1]$ because $C$ is convex. This means that $H^\ast$ restricts to the desired $H$.

Answer 2

The answer is staring at you: the homotopy $f : \mathbb R \times [0,1]$ is defined by the equation $$f(a,t) = a \cdot t + x_0 \cdot (1 - t) $$ This is obviously continuous on the entire domain $\mathbb R \times [0,1]$, and obviously $f(a,0)=x_0$, $f(a,1)=a$.

Now, perhaps you might say "But that's backwards!", because perhaps what you really want is $f(a,0)=a$ and $f(a,1)=x_0$. If so, then your formula is easily tweaked: $$f(a,t) = a \cdot (1-t) + x_0 \cdot t $$