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This is a homework question and I would appreciate if someone could check my reasoning. This seems too simple to be true.

Proof: Assume there exists $X$ with a non-constant function $f:S^n\rightarrow X$ such that $f\not\simeq c_x$. This implies that $1 = |\pi_0(S^n)|\neq|\pi_0(X)| \implies |\pi_0(X)| \geq 2 \implies X$ is non-contractible.

average math enjoyer
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  • From How to ask a good question: "Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question." – jjagmath May 02 '23 at 00:40
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    I can’t make heads or tails of what you’ve said. Are you going to exhibit an $X$ or prove none can exist? If the latter, a proof must suppose $X$ exists with the homotopy conditions and show it must be contractible. What are you doing? – Ted Shifrin May 02 '23 at 00:50
  • Do you mean like $X=S^0$? – John Palmieri May 02 '23 at 00:53
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    The condition does not imply $|\pi_0(X)| \neq 1$. – Qi Zhu May 02 '23 at 05:22
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    Every totally disconnected space has your property. But none of them is contractible, except for single point. In particular every discrete space with at least two points is a counterexample. – freakish May 02 '23 at 05:54

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Your proof does not make sense. What you try to show is that if there exists $f : S^n \to X$ which is not homotopic to a constant function, then $X$ is not contractible. This is trivial: If $X$ is contractible, then all $\phi : Y \to X$ are homotopic to a constant function. In your "proof" you claim that $X$ is not path-connected (i.e. $\pi_0(X)$ has more than one element), but this is wrong. The identity map on $S^n$ is not null-homotopic, but $S^n$ is path-connected for $n > 0$.

The Warsaw circle $W$ is an example of a compact metrizable space which is not contractible, but having all $\pi_i(W) = 0$.

See

Paul Frost
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