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.



Do Smash Products and Quotients Commute?
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 $X = \*$).


