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Just to ask a quick question regarding a corollary 4.12 in Hatcher: "A CW pair $(X,A)$ is n-connected if all the cells in $X-A$ have dimension greater than $n$. In particularly the pair $(X,X^n)$ is n-connected, hence the inclusions $X^n\hookrightarrow X$ induces isomorphism on $\pi_i$ for $i<n$.

Is the reason because $\pi_i(X)$ are all 0, thus the maps must be (trivially) isomorphisms? This is the conclusion I got after checking the long exact sequence, just to confirm if the reason is indeed so.

Long exact sequence of the pair $(X,X^n)$: $\dots\to\pi_n(X^n,x_0)\xrightarrow{i_*}\pi_n(X,x_0)\xrightarrow{j_*}\pi_n(X,X^n,x_0)\xrightarrow{\partial}\pi_{n-1}(X^n,x_0)\to\dots\to\pi_0(X,x_0)$

Since $(X,X^n)$ is n-connected, what I thought is since $\pi_n(X,X^n,x_0)=0$, anything after it is also 0, thus the maps are trivially isomorphisms.

yoyostein
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    I do not at all understand your logic. Take, say, $X=S^1 \times S^2$, $n=2$. Then neither $X$ nor $X^2$ are simply connected. –  Jan 02 '16 at 13:04
  • @MikeMiller What I deduced is since $(X,X^n)$ is n-connected, $\pi_n(X,X^n,x_0)=0$. Thus, applying the boundary map $\partial$ leads to $\pi_{n-1}(X^n,x_0)=0$. And so on, all the later terms are zero? – yoyostein Jan 03 '16 at 02:51
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    $n$-connected by definition means $\pi_i(X,A)=0$ for all $i \leq n$. You don't need to prove these relative groups are zero. I think I originally misread your post: you were just saying the relative groups are automatically zero, not the non-relative groups $\pi_i(X)$, right? Because I agree with that. –  Jan 03 '16 at 02:53
  • @MikeMiller Yes, I was saying the relative groups were automatically zero (due to the n-connectedness). – yoyostein Jan 03 '16 at 04:45
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    My apologies for the misreading. –  Jan 03 '16 at 04:45
  • @MikeMiller No problem, thanks for your help – yoyostein Jan 03 '16 at 04:46

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