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Let $(X,\tau)$ and $(Y,\sigma)$ be topological spaces and let $(X\times Y,\tau\times\sigma)$ be the space with the box topology. Since I never heard of it I guess that there is no space $X\otimes Y$ with a bicontinuous (that is separately continuous here, not a homeomorphism) map $\mu$ such that if $f$ is bicontinuous then there exist an unique continuous map $\varphi$ making the diagram commute. $\require{AMScd}$ \begin{CD} X\times Y @>\mu>> X\otimes Y\\ @V f V V\# @VV \exists!\varphi V\\ Z @= Z \end{CD} How to prove that such a $X\otimes Y$ in general not exists?

Daniel Fischer
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Lehs
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  • What does "(bi)" in "(bi)continuous" mean? I don't understand what you're aiming at, $X\times Y$ (with the product topology, which however for finitely many factors coincides with the box topology) is already the solution to a universal problem. In what direction do you want to step beyond that? – Daniel Fischer Nov 29 '14 at 22:29
  • I'm not sure what you mean by $f$ and $\mu$ being "bicontinuous". Usually it seems to mean a homeomorphism, which would make the existence of $\phi$ trivial. – Hew Wolff Nov 29 '14 at 22:31
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    I am pretty sure that $X \times Y \to Z$ is called bi-continuous iff for every $x \in X$ the resulting map $Y \to Z$ is continuous, and similarly for the other variable. – Martin Brandenburg Nov 29 '14 at 22:39
  • I see, so we are looking for a map which is continuous in each variable separately, and which is initial for this condition. This would be analogous to the tensor product of (say) abelian groups, which is an initial bilinear map from the product group. – Hew Wolff Nov 29 '14 at 22:47
  • @MartinBrandenburg Aha. I was pretty sure that property is usually called "separately continuous". – Daniel Fischer Nov 29 '14 at 22:48

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The general notion of a "bimorphism" and the classifying tensor product has been studied in the paper

B. Banaschewski and E. Nelson. Tensor products and bimorphisms. Canad. Math. Bull, 19(4):385-402, 1976.

It contains a general theorem which shows that tensor products "almost always" exist. In the case of topological spaces, the construction looks as follows:

First, consider the coproduct $P=\coprod_{x \in X} Y \sqcup \coprod_{y \in Y} X$. For each $x \in X$ we have an inclusion $i_x : Y \to P$. For each $y \in Y$ we have an inclusion $j_y : X \to P$. Define $X \otimes Y$ to be the quotient space $P/{\sim}$ where $i_x(y) \sim j_y(x)$. This obviously satisfies the required universal property. Notice that the underlying set of $X \otimes Y$ is just $\{(x,y) : x \in X, y \in Y\}$, and the topology looks as follows: A subset $U$ is open iff for every $x \in X$ the set $\{y \in Y : (x,y) \in U\}$ is open in $Y$ and for every $y \in Y$ the set $\{x \in X : (x,y) \in U\}$ is open in $X$.

If we apply this type of construction to the category of modules, we also obtain an alternative construction of the tensor product of modules (SE/291644).

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Martin Brandenburg
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  • Thanks for the bibitem-tip! Now I will be able to formulate tensor products for the constructs in my awful project. But the tensor product topology above seems quite like the box topology? – Lehs Nov 30 '14 at 00:27
  • I don't know why you refer to "box topology"; since we have two factors the box topology is just the usual product topology. The topology on $X \otimes Y$ usually has much more open subsets (because there are many separately continuous functions which are not continuous). – Martin Brandenburg Nov 30 '14 at 00:31
  • OK! Thank you for your help! – Lehs Nov 30 '14 at 00:33
  • Link to the paper : http://cms.math.ca/openaccess/cmb/v19/cmb1976v19.0385-0402.pdf – Noix07 Oct 29 '15 at 15:57