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I want to show the following:

$X$ $n$-connected $\iff $ any continuous map $f:K \rightarrow X$ where $K$ is a cell complex of dimension $\leq n$ is homotopic to a constant map

For this I think I can use the following: $X$ $n$-connected $\iff $ every continuous map $f: S^n \rightarrow X$ is homotopic to a constant map.

Proof:

"$\Leftarrow$"

If any continuous map $f:K \rightarrow X$ where $K$ is a cell complex of dimension $\leq n$ is homotopic to a constant map then any $f: S^n \rightarrow X$ is homotopic to a constant map. So $X$ is $n$-connected.

"$\Rightarrow$"

I'm not sure how to proceed in this direction. I know $X$ is $n$-connected and so $\pi_i (X) = 0$ for all $i \leq n$. I also know any $f: S^i \rightarrow X$ is null-homotopic.

How to proceed from here? Many thanks for your help!

Rudy the Reindeer
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    Hint: If $X$ is (h.e. to) a CW-complex, then you can replace $f$ by a cellular map (cellular approximation). Then, since $K$ is a cell complex of dimension $\leq n$, it is built inductively by attaching one cell (of dim. $\leq n$) at a time. – Shaun Ault Sep 07 '11 at 12:03
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    I have to confess I sometimes find the way mathematicians write "any" when they mean "every" irritating. "If any continuous map is XXXXX then YYYYYYY" could be taken by perfectly reasonable readers to mean "If there is any continuous map that is XXXXX, then YYYYYYY." Or it could mean "If it is the case that any continuous map, no matter which one, is XXXXX, then YYYYYYY", and that's quite a different thing. "Every" admits no such ambiguity. – Michael Hardy Sep 07 '11 at 13:53
  • @Michael: See also John Milne's Mathlish page (scroll down to Be careful with your use of "any"). – t.b. Sep 07 '11 at 16:39
  • @Michael: I thought that too, several times even. But then I always thought I'm just not advanced enough yet... – Rudy the Reindeer Sep 08 '11 at 07:30
  • @Shaun Ault: But $X$ is not necessarily a CW complex. It's just a topological space. – Rudy the Reindeer Sep 08 '11 at 08:55

2 Answers2

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This is an application of the second theorem of chapter 10.3 in May: A Concise Course in Algebraic Topology, i.e. there you can find your proof.

AlexE
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  • Really? The theorem works for $\dim{K} < n$ but we have given $\dim{K} \leq n$? Maybe I'm missings something here... – Qi Zhu Apr 28 '23 at 19:14
  • If $X$ is $n$-connected, then the map $X \to $ to a single point is an $(n+1)$-equivalence (in the terminology of May). By the cited theorem of May this implies that the induced map $[K,X] \to [K,]$ on homotopy classes of maps is a bijection if $dim(K) < n+1$. – AlexE May 02 '23 at 06:27
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We show that $K^m \hookrightarrow K \rightarrow X$ is nullhomotopic for every $m \leqslant n$ by induction on $m$. This is clear for $m = 0$ since $X$ is path-connected. Suppose the map is nullhomotopic for some $m$ with $0 < m < n$. Since $K^m \hookrightarrow K \rightarrow X$ is nullhomotopic and since CW pairs have the homotopy extension property, there is a homotopy of $K^{m+1} \hookrightarrow K \rightarrow X$ to a map $f \colon K^{m+1} \rightarrow X$ that sends $K^m$ to a points $x_0 \in X$. Hence, $f$ factors through a map $K^{m+1}/K^m \cong \vee \mathbb{S}^{m+1} \rightarrow X$, which is nullhomotopic since $X$ is $n$-connected.

SFSH
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