Maps to Sn homotopic 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.
 A: 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).  
A: Thanks to the condition $f(x) \neq -g(x)$, you can define an explicit homotopy between $f$ and $g$, actually.
The idea is that $S^n$ minus the north pole $N$ deformation retracts onto the south pole $S$. It's possible to write an explicit homotopy $H : S^n \setminus \{N\} \times [0,1] \to S^n \setminus \{N\}$ such that $H(u,0) = u$ and $H(u,1) = S$.
And in fact, $S^n \setminus \{-x\}$ deformation retracts onto $\{x\}$, because if $A_x \in SO(n)$ sends the north pole to $-x$ (and hence sends the south pole to $x$, the rotated homotopy $H$ will be the deformation retraction you want. Therefore you get the D.R. $H_x : S^n \setminus \{-x\} \times [0,1] \to S^n \setminus \{-x\}$, and you can define it in such a way that it continuously depends on $x$ (I'll let you make that precise).
How does that help? Define the new homotopy $G : X \times [0,1] \to S^n$ by
$$G(x,t) = H_{g(x)}(f(x),t)$$
This is well defined precisely because $f(x) \neq -g(x)$. At time $t=0$, this is $f(x)$, and at time $t=1$, this is $g(x)$; and because of the way we define $H_x$, this is continuous. Hence you get a homotopy between $f$ and $g$.
