Let $X$ be a finite CW-complex with only even cells $x_1,\ldots, x_k$ and let $Y$ be the complex obtained by attaching one more even cell to $X$, call it $y$. Assume both $X$ and $Y$ are connected. The quotient complex $Y^n/X^n$ has the cell structure with one cell for each product of cells $e_1\times\cdots\times e_n$ where the $e_i$ are either equal to $x_i$ or $y$ and at least one $e_i$ is equal to $y$. I want to say that this amounts to a wedge of products of spheres each product of spheres depending on how many copies of $y$ there is in that cell. Is this correct?


Suppose Y is obtained by attaching a zero-cell, so $Y = X \cup \{\ast\}$. Then $Y^2$ is $$(X \times X) \cup (X \times \{\ast\}) \cup (\{\ast\} \times X) \cup (\{\ast\} \times \{\ast\})$$ and so $Y^2/X^2$ is homeomorphic to $$ \{\ast\} \cup X \cup X \cup \{\ast\}. $$ This can be arbitrarily complicated depending on X.

ADDENDUM: By request, a connected example is the inclusion $\mathbb{CP}^1 \subset \mathbb{CP}^2$. The quotient $Y^2/X^2$, in this case, has a nonzero cohomology operation $Sq^2$ from H6 to H8 with ℤ/2-coefficients, and there is no wedge of products of spheres that can have this cohomology. (You should work out the details.)

  • $\begingroup$ What if I assume that both $X$ and $Y$ are connected? $\endgroup$ – Richard Aug 30 '10 at 17:01
  • $\begingroup$ @Richard: I think you get the same problem if you attach something other than a 0-cell. If $y$ is the new cell, the quotient looks like $\{\ast\}\cup (X\times y)\cup (y\times X)\cup (y\times y)$. $\endgroup$ – Anton Geraschenko Aug 30 '10 at 18:15
  • $\begingroup$ @Anton: Exactly. The sets $(X \times y)$ etc produce suspensions of X sitting inside this quotient space. Unfortunately in order to show that these don't decompose, up to homotopy equivalence, as wedges of products of spheres you need at least a little machinery. $\endgroup$ – Tyler Lawson Aug 30 '10 at 18:44

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