Let $X$ be a set with the discrete topology and let $\pi_n(X)$, as usual, be the set of homotopy classes of maps $f : S^n \rightarrow X$ that map the base point $a$ to the base point $b$. I'm trying to prove that, in this case, $\pi_n(X)$ is singleton for all $n>0$, that is, every two base-point-preserving maps are equivalent. I think I have to show that, given $f$ and $g$ base-point-preserving maps, there is $F:S^n\times[0,1]\rightarrow X$ continuous such that $F(x,0)=f(x)$, $F(x,1)=g(x)$ and $F(a,t)=b$, $\forall \ x\in S^n$ and $\forall \ t\in[0,1]$. But I couldn't notice how to get that using the discrete topology in $X$ and I'd like to get a hint, for kindness. Thank you!

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    $\begingroup$ If $X$ is discrete, then all continuous maps from connected spaces to $X$ have something in common that makes the exercise rather easy. $\endgroup$ – Daniel Fischer Jan 12 '17 at 20:38
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    $\begingroup$ No, if $X$ is discrete, every subset is clopen, and so every continuous map from a connected space to $X$ is constant. So for $n > 0$ there is only one continuous map on $S^n$ that maps the base point to the base point. $\endgroup$ – Daniel Fischer Jan 13 '17 at 10:55
  • $\begingroup$ Thanks for your help and sorry for my unfamiliarity with the subject. That is a rather easy exercise, like you said. $\endgroup$ – rgm Jan 13 '17 at 11:00
  • $\begingroup$ When you see the crucial point it's easy. But seeing the crucial point is always a nontrivial part. $\endgroup$ – Daniel Fischer Jan 13 '17 at 11:02

If $X$ is discrete then points are connected components (i.e. $X$ is totally disconnected). In particular if $Y$ is connected then every continuous map $Y\to X$ is constant. Since $S^n$ is connected then the thesis trivially follows since there is only one continuous function $S^n\to X$ that fixes base points.


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