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?



What Is This Quotient Space?
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.)


