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.

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

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 $g$.

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

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).

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.