I've read many times that the smash product of pointed topological spaces is not associative, for example $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not homeomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$. How to prove this?
In general $\mathbb{N} \wedge X = \bigvee_{n>0} X$, so that $\mathbb{N} \wedge (X \wedge Y) = \bigvee_{n>0} (X \wedge Y)$ and $(\mathbb{N} \wedge X) \wedge Y = (\bigvee_{n>0} X) \wedge Y$. By the universal property of the wedge sum, we get a canonical continuous map $$\mathbb{N} \wedge (X \wedge Y) \to (\mathbb{N} \wedge X) \wedge Y,~(n,(x,y)) \mapsto ((n,x),y).$$ It is clearly bijective. For $X=Y=\mathbb{Q}$ it should not be open - why? Probably one has to take some open subset of $\mathbb{Q} \times \mathbb{Q}$ which approaches the base point $(0,0)$ in a weird way?
Can someone explain/supplement Joriki's proof?
The preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ of an open neighbourhood of the origin of $(\mathbb N\wedge\mathbb Q)\wedge\mathbb Q$ must contain a product of open neighbourhoods of the origins of $\mathbb N\wedge\mathbb Q$ and $\mathbb Q$, so its preimage in $\mathbb N\times\mathbb Q\times\mathbb Q$ must contain a set of the form $\{0\}\times\mathbb Q\times U$, where $U\subseteq\mathbb Q$ is a neighbourhood of $0$. However, $\{(n,x,y)\mid|xy|\lt1\}$ doesn't contain such a set, and its image in $\mathbb N\wedge(\mathbb Q\wedge\mathbb Q)$ is open.