At the moment I'm taking an algebraic topology course. In our current exercise we have to prove the following:
Consider any topological space $X$ and any two maps $f, g : X \to S^n$, such that $f(x) \neq -g(x)$. It follows that $f$ is homotopic to $g$.
My idea is to use the mapping cylinders $M_f, M_g$ to construct a homotopy via $$M := M_f \amalg M_g/ (x, 0)_f \sim (x,0)_g$$ But this doesn't seem to work. In addition I don't know how $f(x) \neq -g(x)$ is going to help me (I can see how this could help if $n=1$, but I have no clue how this would work for $n>1$).
I would appreciate any hints towards a solution.
Thanks in advance.
Best Answer
What about this map:
$$F(x,t)=\dfrac{tf(x)+(1-t)g(x)}{\|tf(x)+(1-t)g(x)\|}$$ Remark: for all $t\in[0,1]$ , $tf(x)+(1-t)g(x)\neq 0$. If $tf(x)+(1-t)g(x)=0$ then $tf(x)=-(1-t)g(x)$ applying $\|.\|$ we get $t=1-t$ hence $t=1/2$, thus $f(x)=-g(x)$ (this is not possible).