Let $X$ be a subcomplex of a CW-complex $Y$. Is $(Y/X)^{\wedge k}$ homotopy equivalent to $Y^{\wedge k}/X^{\wedge k}$, where $\wedge k$ is the $k$-fold smash product? I know it is not true for products but am having a hard time visualizing for smash products.
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The easiest way I know to say what is going on is to resort to looking at "products" of pairs: $$ (X, A) \times (Y, B) = ( X\times Y , A\times Y \cup X\times B). $$ The point of this notation is that the functor $(X, A) \mapsto (X/A, *)$ carries $(X, A) \times (Y, B)$ to $X/A \wedge Y/B$. We can iterate this procedure, and I'll write $T^n(Y,X)$ for the subspace of $Y^n$ satisfying $$ (Y, X)^n = ( Y^n, T^n(Y, X)). $$ Thus $(Y/X)^{\wedge n} = Y^n /T^n(Y,X)$. You can easily check that $$ T^n( Y, X) = \lbrace (y_1, \ldots, y_n) \mid y_i \in X\ \mbox{for at least one $i$}\rbrace. $$ On the other hand $Y^{\wedge n}/X^{\wedge n}$ is the quotient of $Y^n$ by the subspace $$ T^n(Y,*) \cup X^n, $$ which is different (unless $X = *$). |
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$X\wedge X\subset (Y\wedge X)\cup (X\wedge Y)\subset Y\wedge Y$, with quotients $((Y/X)\wedge X)\vee (X\wedge (Y/X))$ and $(Y/X)\wedge (Y/X)$. |
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