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 $U$?
Many thanks in advance!
$\Bbb R^2{/}((\Bbb R \times \{0\}) \cup (\{0\}\times \Bbb R))$ is not locally compact at the "identified point" $0 \land 0$. So even false for nice manifolds etc.