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?