most recent 30 from http://mathoverflow.net 2013-05-25T13:38:42Z http://mathoverflow.net/feeds/question/36810 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36810/do-smash-products-and-quotients-commute Richard 2010-08-26T20:33:03Z 2010-10-25T03:57:32Z

<p>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.</p>

http://mathoverflow.net/questions/36810/do-smash-products-and-quotients-commute/43442#43442 Jeff Strom 2010-10-24T22:33:07Z 2010-10-25T03:17:33Z

<p>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 $(Y/X)^{\wedge n} = Y^n /T^n(Y,X)$. </p> <p>You can easily check that $$ T^n( Y, X) = \lbrace (y_1, \ldots, y_n) \mid y_i \in X\ \mbox{for at least one $i$}\rbrace. $$</p> <p>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 $X = *$).</p>

http://mathoverflow.net/questions/36810/do-smash-products-and-quotients-commute/43469#43469 Tom Goodwillie 2010-10-25T03:57:32Z 2010-10-25T03:57:32Z

<p>$X\wedge X\subset (Y\wedge X)\cup (X\wedge Y)\subset Y\wedge Y$, with quotients $((Y/X)\wedge X)\vee (X\wedge (Y/X))$ and $(Y/X)\wedge (Y/X)$.</p>