This is a topology question:
Let $X$ be a space. Prove (with explicit formulas) that $f \simeq 0: X → \mathbb R$ for any map $f : X → \mathbb R$. Here, $\simeq$ stands for homotopic, and $\mathbb R$ is the set of the real numbers.
These are my thoughts:
Suppose $X$ is a topological and $f: X → \mathbb R$ is continuous. Define $H$:$$H: X \times [0,1] → \mathbb R, \quad H(x,t) = tf(x),\quad t \in [0,1], \ x \in X$$
Then $H$ is continuous on $X \times [0,1]$ with $H(x,0)=0$ and $H(x,1)=f(x)$ for all $x \in X$. Thus, $X$ is a homotopy between $0$ and $f$, thus $f \simeq 0$.
Am I correct? If not, could you provide the correct way to solve this? Any help is appreciated!
Use the straight line homotopy:
$H(x,t) = (1-t)0+t f(x) = f(x)t$
Which can be done if $f(x)t \in \mathbb{R}$ which it does.