# Associativty of smash product on compact spaces

Let $Y,Z$ be compact or $X,Z$ locally compact. Then the canonical bijection $$(X \wedge Y) \wedge Z \rightarrow X \wedge (Y \wedge Z)$$ is a homeomoprhism.

I can prove the case when $X,Z$ are locally compact using exponential law. But I don't see how one approaches the case when $Y,Z$ compact. Hints?

It is clear that we have surjections

$$p_{X;Y,Z} : X \times Y \times Z \stackrel{id_X \times q_{Y,Z}}{\longrightarrow} X \times (Y \wedge Z) \stackrel{q_{X,Y \wedge Z}}{\longrightarrow} X \wedge (Y \wedge Z)$$

$$p_{X,Y,Z} : X \times Y \times Z \stackrel{q_{X,Y} \times id_Z}{\longrightarrow} (X \wedge Y) \times Z \stackrel{q_{X \wedge Y,Z}}{\longrightarrow} (X \wedge Y) \wedge Z$$

where the $q$-maps are quotient maps. Moreover, there is a bijection $\tau : X \wedge (Y \wedge Z) \to (X \wedge Y) \wedge Z$ such that $\tau \circ p_{X;Y,Z} = p_{X,Y;Z}$.

It therefore suffices to show that both $p$-maps are quotient maps. In turn it suffices to show that $id_X \times q_{Y,Z}$ and $q_{X,Y} \times id_Z$ are quotient maps.

If $X,Z$ are locally compact, this is well-known. It remains to show that if $Y,Z$ are compact, then $id_X \times q_{Y,Z}$ is a quotient map. In this case we have $q_{Y,Z} : Y \times Z \to Y \wedge Z = (Y \times Z)/(Y \vee Z)$ where $Y \vee Z$ is compact. Now apply exercise 16 in section 2.2 in the book "Algebraic Topology" by tom Dieck.

Frankly, I have not done the exercise, but it seems to be straightforward.

• Thanks a lot Paul, here is my attempt of proof of statement: let $U \subseteq X/A \times Y$, such that $(q \times id)^{-1}(U)$ is open. Then if $(x,y) \in U$, $A \times \{ y \} \subseteq (q \times id)^{-1}(U)$. By tube lemma, exists open $U_y \times V_y \subseteq (q \times id)^{-1}(U)$, hence $q(U_y) \times V_y$ is our desired open set containing $(x,y)$, since $q$ itself is a quotient map. – CL. Aug 31 '18 at 14:35
• However, I don't see why $q_{X,Y} \times id_Z$ is a quotient map. – CL. Aug 31 '18 at 14:35
• @CyrylL. It is a quotient map because $Z$ is compact – Paul Frost Aug 31 '18 at 17:15
• I thought this only holds for $Z$ locally compact, and $Z$ compact does not imply locally compact unless Hausdorff(?) – CL. Aug 31 '18 at 17:24
• @CyrylL. I see. In my interpretation "compact" and "locally compact" include "Hausdorff". Without requiring "Hausdorff" both concecpts are not really powerful. However, I am aware that this is not a commonly accepted standard. If you prefer to work without "Hausdorff", you are in accordance with tom Dieck and you cannot conclude that compact implies locally compact. You should then try to check whether Theorem (2.4.6) is also true for compact $Z$. If that fails, then either $q_{X,Y} \times id_Z$ is in general nor a quotient map or it requires an individual proof. – Paul Frost Aug 31 '18 at 17:50