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Let $Y$ be a connected CW-complex and $X\subset Y$ a connected CW-subcomplex. Suppose that each cell of $X$ is the boundary of a cell of $Y$. Is this enough to conclude that $X$ is contractible in $Y$ (the inclusion map is homotopic to a constant map)? If the answer is no, then which condition could be enough to get the contractibility?

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Since $X\subset Y$ this just means that every cell of $X$ is the boundary of some other cell, which could also be in $X$.Is that what you had in mind? – Jim Conant Jan 28 '11 at 15:08
Yes. But the cells bounded by cells in $X$ are possibly cels of $Y$ and not of $X$. – user12570 Jan 31 '11 at 8:55
up vote 4 down vote accepted

If you mean that every cell in $X$ is the image of the boundary of a cell in $Y$ then the answer is no -- consider for example the standard inclusion $S^1 = RP^1 \hookrightarrow RP^2$. The boundary of the $2$-cell is $RP^1$ (but it runs around twice, so $\pi_1 (RP^2) = \mathbb{Z}/2$ and the $1$-cell represents the generator).

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Yes, you right. But can you add some mild condition to get the contractibility? – user12570 Jan 31 '11 at 9:56

The condition "each cell of $X$ is the boundary of a cell in $Y$" is very strong.

In particular, it implies that each cell of $X$ is attached to the rest of $X$ by a constant attaching map (otherwise, it would have no chance of being the boundary of something else)

Thus, $X$ is a wedge of spheres.

I think that, at this point, it is easy to see how to finish the argument and answer the question in the affirmative: yes $X$ is contractible in $Y$.

I feel however, that the question asked maybe wasn't the one intended.

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I regard "each cell of $X$ is the boundary of a cell of $Y$" as ambiguously formulated. – John Klein Jan 28 '11 at 14:09
I don't think it's ambiguous, but I think its (unique, as far as I can see) meaning, the one André Henriques used, might well not be what the OP intended. – Andreas Blass Jul 12 '11 at 14:46

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