Homotopic functions

Question

Let $X$ be a space and $f,g:X\rightarrow S^n$, s.t $f(x)\neq g(x)$ ( $f,g$ continuous) for all $x$. Show that they are homotopic.

Can you give me any hints on how to start attacking this exercise? thanks

Answer 1

This is not true. If n is even the map $x \rightarrow -x$ from $S^n$ to itself and the identity map satisfy the hypothesis of your post, but the degree of the first is $-1$ while the second has degree $1$ which means they are not homotopic.

If you want a brutally explicit counterexample let $n=0$ then there are only two possible maps that satisfy the hypothesis if $X$ is connected, the constant maps to the two points. Clearly they are not homotopic.

Answer 2

Construct $H(x,t): X \to S^n, = \frac{tf(x) + (1-t)g(x)}{||tf(x) + (1-t)g(x) ||}$. Clearly $H$ is continuous since the denominator is always nonzero (by assumption $f \neq g$), and $H(x, 0)=g(x)$ and $H(x,1) = f(x)$.