If $X,Y,Z$ are locally compact Hausdorff spaces then
(A) $X\wedge (Y\wedge Z)$ and $(X\wedge Y)\wedge Z$ are homeomorphic.
(B) $X\wedge Y$ and $Y\wedge X$ are homeomorphic.
Can anybody prove or give some hint to prove it. I am not getting any idea how to proceed.
(A) As Tyrone comments, you have to use the well-known fact that if $p : A \to B$ is a quotient map and $C$ is locally compact Hausdorff, then also $p \times 1_C$ and $1_C \times p$ are quotient maps.
Note that $(A,a_0) \wedge (B,b_0) = (A \times B)/(A \times \{b_0\} \cup \{a_0\} \times B)$ with basepoint $*$ = common equivalence class of the points in $A \times \{b_0\} \cup \{a_0\} \times B$. If $p_{A,B} : A \times B \to (A,a_0) \wedge (B,b_0)$ denotes the quotient map, then $p^{-1}_{AB}(*) = A \times \{b_0\} \cup \{a_0\} \times B$ and $p^{-1}_{A,B}([a,b]) = (a,b)$ for $[a,b] \ne *$.
Thus we get quotient maps $$\phi : X \times Y \times Z \stackrel{p_{X,Y} \times 1_Z}{\longrightarrow} (X \wedge Y) \times Z \stackrel{p_{X \wedge Y,Z} }{\longrightarrow} (X \wedge Y) \wedge Z ,$$ $$\psi : X \times Y \times Z \stackrel{1_X \times p_{Y,Z}}{\longrightarrow} X \times (Y \wedge Z) Z \stackrel{p_{X, Y \wedge Z} }{\longrightarrow} X \wedge (Y \wedge Z) .$$ Let us check what these maps do. We have $$\phi^{-1}([[x,y],z]) = \begin{cases} X \times Y \times \{z_0\} \cup X \times \{y_0\} \times Z \cup \{x_0\} \times Y \times Z & [[x,y],z] = * \\ (x,y,z) & [[x,y],z] \ne * \end{cases} $$ A analogous formula is obtained for $\psi$. Thus both maps make the same identifications in $X \times Y \times Z$ which shows that the quotient spaces are homeomorphic. In fact, there exists a unique homeomorphisms $h : (X \wedge Y) \wedge Z \to X \wedge (Y \wedge Z)$ such that $h \circ \phi = \psi$. It is given by $h(([[x,y],z]) = [x,[y,z]]$.
(B) This is almost trivial. Let $\tau = \tau_{X,Y} : X \times Y \to Y \times X, \tau(x,y) = (y,x)$, be the switch map. It induces a unique function $\tau' : X \wedge Y \to Y \wedge X, \tau'([x,y]) =[y,x]$. We have $\tau' \circ p_{X,Y} = p_{Y,X} \circ \tau$ which shows that $\tau'$ is continous. But $\tau'_{Y,X} \circ \tau'_{X,Y} = 1$, thus $\tau'$ is a homeomorphism.
For based spaces $K,L$ Let $K\wedge L=K\times L/K\vee L$ be their smash product and $q_{K,L}:K\times L\rightarrow K\wedge L$ the quotient map.
Now fix based spaces $X,Y,Z$. We will assume that both $X$ and $Z$ are locally compact. Both $X\wedge(Y\wedge Z)$ and $(X\wedge Y)\wedge Z$ are quotients of $X\times Y\times Z$. Consider the two compositions $$\pi_1:X\times Y\times Z\xrightarrow{1\times q_{Y,Z}} X\times (Y\wedge Z)\xrightarrow{q_{X,Y\wedge Z}} X\wedge (Y\wedge Z)$$ and $$\pi_2:X\times Y\times Z\xrightarrow{q_{X,Y}\times 1} (X\wedge Y)\times Z\xrightarrow{q_{X,Y\wedge Z}} (X\wedge Y)\wedge Z$$ Since $X$ is locally compact, $1\times q_{Y,Z}$ is a quotient map, and thus $\pi_1$ is a quotient map. See here for some discussion of this fact. Thus the map $\pi_2$ factors over the quotient and induces a continuous bijection $$\theta_1:X\wedge (Y\wedge Z)\rightarrow (X\wedge Y)\wedge Z.$$ On the other hand, since $Z$ is locally compact, $q_{X,Y}\times 1$, and hence $\pi_2$ are quotient maps. Thus $\pi_1$ factors to induce a continuous bijection $$\theta_2:(X\wedge Y)\wedge Z\rightarrow X\wedge (Y\wedge Z).$$ Then $\theta_1,\theta_2$ are inverse homeomorphisms.
Note that in general the assumption of local compactness cannot be dropped. The rationals are not locally compact and the spaces $(\mathbb{Q}\wedge\mathbb{Q})\wedge\mathbb{Z}$ and $\mathbb{Q}\wedge(\mathbb{Q}\wedge\mathbb{Z})$ are not homeomorphic.
