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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.

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  • $\begingroup$ Of course $f, g$ are continuous. $\endgroup$ – Lucas Boucke Sep 26 '14 at 9:18
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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 $g$.

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  • $\begingroup$ Thank you for this explanation! $\endgroup$ – Lucas Boucke Sep 27 '14 at 12:28
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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).

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    $\begingroup$ That's probably the easiest solution (but I think Najibs answer is more interesting because it gives more insight into why the statement is true). Your answer shows how focused I was on that arbitrary space $X$ - I assumed that only a very abstract prive could do the job... - Thanks for your answer ; ) $\endgroup$ – Lucas Boucke Sep 29 '14 at 23:23

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