Is the tensor product of compactly generated Hausdorff abelian groups again Hausdorff? Consider the tensor product $G \otimes_{\mathbb{Z}} H$ of two abelian groups $G$ and $H$. If $G$ and $H$ are topological groups, we can give $G \otimes_{\mathbb{Z}} H$ a topology as follows. For any $k$, consider the function.
\begin{align}
\phi_k: G^k \times H^k = G \times \dots \times G \times H \times \dots \times H &\to G \otimes_{\mathbb{Z}} H, \\
(g_1, \dots, g_k, h_1, \dots, h_k) &\mapsto \sum_{i=1}^k g_i \otimes h_i.
\end{align}
We topologize $G \otimes_{\mathbb{Z}} H$ using the finest topology making all functions $\phi_k$ continuous. It follows that $G \otimes_{\mathbb{Z}} H$ has the universal property with respect to all continuous bilinear maps $G \times H \to G'$ into a topological group $G'$. 
The question:


*

*If we assume that $G$ and $H$ are compactly generated Hausdorff topological spaces, is $G \otimes_{\mathbb{Z}} H$ again compactly generated Hausdorff?


(It is automatically compactly generated, being a quotient of the compactly generated space $\bigsqcup_{k=1}^\infty G^k \times H^k$. The problem is Hausdorffness.)


*If the above is not true, we could replace $G \otimes_{\mathbb{Z}} H$ by its Hausdorff quotient. Is it true that the resulting functor $(G,H)\mapsto {\rm Haus}(G \otimes_{\mathbb{Z}} H)$ is universal with respect to bilinear maps $G \times H \to G'$ into a compactly generated Hausdorff topological abelian group $G'$? In particular, is this functor naturally associative?

 A: This is not true. Our example will be $G = H = S^1$ with its standard topology.
Let $\mu \subset S^1$ be the subgroup of roots of unity. The group $S^1$ is divisible and $\mu$ consists entirely of torsion, and so
$$
\mu \otimes S^1 = 0.
$$
This lets us conclude that the natural map of groups
$$
S^1 \otimes S^1 \to (S^1/\mu) \otimes (S^1/\mu)
$$
is an isomorphism. (Here $S^1/\mu$ is the group theoretic quotient, not the set-theoretic one.)
When we bring topologies in, the quotient group $S^1/\mu$ has a natural quotient topology: this topology is the indiscrete topology because $\mu$ is dense in $S^1$. Similarly, the quotient map $S^1 \times S^1 \to (S^1/\mu) \times (S^1/\mu)$ makes the target indiscrete, and so in this case the product of the quotient topologies is the quotient of the product topologies.
The universal property of the topologized tensor product is that a topological group homomorphism $S^1 \otimes S^1 \to G$ is the same as a continuous map $S^1 \times S^1 \to G$ that is a group homomorphism in each variable; this is the same as a continuous map $S^1/\mu \times S^1/\mu \to G$ that is a group homomorphism in each variable and which pulls back to a continuous map on $S^1 \times S^1$; because of the above quotient topology property, that's the same as a map of topological groups $(S^1/\mu) \otimes (S^1/\mu) \to G$. So $S^1 \otimes S^1$ coincides with $(S^1/\mu) \otimes (S^1/\mu)$ as a topological group.
We now have to show that $S^1/\mu \otimes S^1/\mu$ is not Hausdorff. It's not immediate that it has the indiscrete topology and so we need to be a little careful, but the map $x \mapsto x \otimes \pi$ does determine a continuous 1-to-1 map from $S^1/\mu$ into the tensor product, and so the target can't possibly be Hausdorff.
