# Smash product of locally compact spaces

Is it true that the smash product $$X\wedge Y$$ of two locally compact pointed spaces $$X$$ and $$Y$$ is again locally compact?

My attempt to prove it:

The smash product $$X\wedge Y$$ is defined as the quotient space $$X\wedge Y = (X\times Y) / (X\times \{y_0\} \cup \{x_0\}\times Y)$$ where $$x_0$$ and $$y_0$$ are the basepoints of $$X$$ and $$Y$$, respectively. We denote by $$p: X\times Y \to X\wedge Y$$ the canonical quotient map and write $$x\wedge y$$ for the element in $$X\wedge Y$$ represented by $$(x,y)\in X\times Y$$.

A space $$Z$$ is called locally compact if for every point $$z\in Z$$ and every open neighbourhood $$U\subset Z$$ of $$z$$, there exists a compact neighbourhood $$K\subset Z$$ of $$z$$ with $$K\subset U$$.

Let $$x\in X$$ and $$y\in Y$$ be any points, and let $$U\subset X\wedge Y$$ be an open neighbourhood of $$x\wedge y$$. Then $$p^{-1}(U)\subset X\times Y$$ is an open neighbourhood of $$(x,y)$$, so there exist open subsets $$V\subset X$$ and $$W\subset Y$$ with $$(x,y) \in V\times W\subset p^{-1}(U).$$ Since $$X$$ and $$Y$$ are locally compact, there exist compact neighbourhoods $$K\subset X$$ of $$x$$ and $$L\subset Y$$ of $$y$$ with $$K\subset V$$ and $$L\subset W$$. Now, $$p(K\times L)$$ is a compact subspace of $$X\wedge Y$$ containing $$x\wedge y$$, and we have $$p(K\times L) \subset p(V\times W) \subset U.$$ However, $$p$$ does not in general preserve neighbourhoods, so we do not know whether $$p(K\times L)$$ is still a neighbourhood of $$x\wedge y$$ in $$X\wedge Y$$:

Let $$V'\subset X$$ and $$W'\subset Y$$ be open subsets with $$(x,y)\in V'\times W'\subset K\times L$$. If $$x_0\notin V'$$ and $$y_0\notin W'$$, or if $$X\times\{y_0\}\cup\{x_0\}\times Y$$ is entirely containted in $$V'\times W'$$, then $$V'\times W'$$ is saturated with respect to $$p$$, so in this case, $$p(V'\times W')$$ is open in $$X\wedge Y$$ and, thus, $$K\times L$$ is a (compact) neighbourhood of $$x\wedge y$$ in $$X\wedge Y$$. But what if the intersection of $$V'\times W'$$ with $$X\times\{y_0\}\cup\{x_0\}\times Y$$ is neither empty nor all of the latter space? Does $$x\wedge y$$ still has a compact neighbourhood in $$X\wedge Y$$ contained in $$U$$?

• It seems unclear to me whether $x_0 \land y_0$ has any compact neighbourhood in general. I wouldn’t expect so. I think it’s already false for two copies of the reals with $0$ as point e.g. Commented Oct 31, 2021 at 9:22
• What about the reduced suspension $X\wedge S^1$? Is this locally compact if $X$ is locally compact? (Now, $Y=S^1$ is even compact Hausdorff.) Commented Oct 31, 2021 at 11:32
$$\Bbb R^2{/}((\Bbb R \times \{0\}) \cup (\{0\}\times \Bbb R))$$ is not locally compact at the "identified point" $$0 \land 0$$. So even false for nice manifolds etc.