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I am trying to learn Cellular Homology from Allen Hatcher's AT book, but stuck in first Lemma (2.34) itself. While introducing Cellular Homology Hatcher in his AT, Section 2.2 Lemma 2.34 says $X^n/X^{n-1}$ is a wedge sum of $n$ - spheres, one for each $n$- cell of $X$. Where $X$ is a $CW$ Complex.

I am unable to prove this, couldn't even think a bit how to start or where to start. Please guide me.

Ram
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    It sounds like you're learning from Allen Hatcher's book, not the person himself. Regarding your question, you build $X^n$ from $X^{n-1}$ by attaching $n$-dimensional discs along maps of their boundary spheres. So if you crush $X^{n-1}$, you are crushing not only all the boundary spheres to one common point, but everything else in the space too. – Ryan Budney Aug 23 '13 at 21:46
  • @RyanBudney, sorry for my English, thanks for your comment. – Ram Aug 23 '13 at 21:54

1 Answers1

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Each $n$-cell of $X$ is a ball glued along its boundary to the $(n-1)$-skeleton. If you collapse the $(n-1)$-skeleton to a point, you're collapsing the boundary of each $n$-cell to a point and identifying all those points. A ball whose boundary is collapsed to a point is a sphere, so $X^n/X^{n-1}$ is a sphere for each $n$-cell, all glued along the point to which the $(n-1)$-skeleton was collapsed.

Neal
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    Neal, thanks for your answer, your answer gave me better understanding of CW Complexes. – Ram Aug 23 '13 at 21:57