# Do Smash Products and Quotients Commute?

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.

• $Y = \{\ast,1,2\}$, $X = \{\ast,1\}$, $k=2$. – Tyler Lawson Aug 26 '10 at 21:06
• I too find smash products hard to visualize, but tyler's example is great cus you can visualize it. Also, smash products will not behave like the ordinary product because they are not a categorical product. I believe the reason you get that commutativity is because the products satisfy a universal property. Also, once you start talking about smash products you will probably want to learn about cofibers and then ask the question. – Sean Tilson Aug 26 '10 at 21:26
• @Richard: Are you sure about the products? What happens if we have CW-complexes which are not locally compact? At least it works if we work in the category CGHaus. – Martin Brandenburg Oct 24 '10 at 23:14
• @Sean: Are you sure? I thought smash products were the product in the category of pointed spaces... since if we have pointed maps into X and Y we certainly get a pointed map into X smash Y, right? But I guess I haven't thought about uniqueness and continuity – Dylan Wilson Oct 25 '10 at 4:46
• @Dylan Wilson: products in pointed spaces are just Cartesian products $(X,x) \times (Y,y) = (X \times Y, (x,y))$. That obviously has the right universal property. The smash product is a nice monoidal structure basically because (working in a convenient category of spaces) it is left adjoint to forming pointed mapping spaces (so it is the "tensor product" corresponding to the "internal hom" in pointed spaces). – Omar Antolín-Camarena Mar 29 '12 at 22:47

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 $X = *$).
$X\wedge X\subset (Y\wedge X)\cup (X\wedge Y)\subset Y\wedge Y$, with quotients $((Y/X)\wedge X)\vee (X\wedge (Y/X))$ and $(Y/X)\wedge (Y/X)$.