The n-Category Café

Hi Tom –

I ran across this query of yours from two and a half years ago (!), and decided to give it another crack. You wrote

[Any] Hausdorff non-locally-compact space $Y$, such as $\mathbb{Q}$, has the property that $− \times Y$ doesn’t preserve quotient maps. And quotient maps are coequalizers, so this confirms my suspicions. Of course, this still doesn’t give us a particular coequalizer/quotient that fails to be preserved by − \times $\mathbb{Q}$. Maybe a close reading of Day and Kelly would yield one…

I never did get hold of Day & Kelly, but I think what I wrote earlier can be repackaged to yield a specific example. Not an easy example unfortunately – I wish I knew one.

Let $Y$ be Hausdorff but not locally compact. To give a coequalizer not preserved by the functor $Y \times -$, it suffices to give an example of a colimit not preserved by this functor. (For we can present the colimit as a coequalizer of maps between coproducts, and since $Y \times -$ preserves coproducts, it must not preserve this coequalizer.)

Let $\mathbf{2}$ be Sierpinski space: the set $2 = \{0, 1\}$ where $\{1\}$ is open but $\{0\}$ is not. Let $\mathcal{O}(Y)$ denote the topology of $Y$, and let

$$char: Y \times \mathcal{O}(Y) \to \mathbf{2}$$

be the characteristic function, taking a pair $(y, V)$ to $1$ precisely when $y \in V$. Now consider the diagram in $Top$ whose nodes are all topological space structures on $\mathcal{O}(Y)$ which render $char$ continuous (call such topologies $\mathcal{T}$ on $\mathcal{O}(Y)$ strong), and whose arrows are continuous identity functions

$$id: \mathcal{O}(Y)_{\mathcal{T}} \to \mathcal{O}(Y)_{\mathcal{T}'},$$

meaning there is an inclusion $\mathcal{T}' \hookrightarrow \mathcal{T}$ of strong topologies. I claim $Y \times -$ cannot preserve the colimit of this (directed) diagram.

The colimit in $Top$ of this diagram is $\mathcal{O}(Y)$ topologized by the intersection of all strong topologies. If $Y \times -$ preserved this colimit, then we would have an induced continuous map

$$char: Y \times colim_{\mathcal{T}} \mathcal{O}(Y)_{\mathcal{T}} \to \mathbf{2}$$

so that the intersection of all strong topologies would itself be strong. I will show this is impossible, by showing that for every strong topology $\mathcal{T}$, there is a strictly smaller strong topology $\mathcal{T}' \hookrightarrow \mathcal{T}$. This will rely on the non-local-compactness of $Y$. Specifically, we use the fact that if $Y$ is Hausdorff but not locally compact, then we can choose an open $V$ and a point $y \in V$ such that for all open $U$ with $y \in U \subseteq V$, there exists a non-compact closed $C$ with $y \in C \subseteq U$.

A topology $\mathcal{T}$ is strong iff $char$ is continuous iff the elementhood relation $\in_Y = char^{-1}(1)$ is open in $Y \times \mathcal{O}(Y)_{\mathcal{T}}$. In that case, the chosen point $(y, V)$ belongs to $\in_Y$ (i.e., $y \in V$), so there exist open sets $U \subseteq Y$, $D \in \mathcal{T}$, such that

1. $y \in U$
2. $V \in D$
3. $U \times D \subseteq \in_Y$

Lemma: The third condition is equivalent to the condition

$$U \subseteq \bigcap D = \bigcap \{W \in \mathcal{O}(Y): W \in D\}.$$

Together with conditions 1 and 2, this implies $y \in U \subseteq V$.

We can find a non-compact closed set $C \subseteq U$ containing $y$. Let $U_a$ be an open cover of $C$ with no finite subcover. There is no harm in assuming that the difference $V - C$ is one of the $U_a$.

Now comes a trick$^\dagger$ (which I won’t try to motivate). Define a new topology $\mathcal{U}$ to consist of those $D' \subseteq \mathcal{O}(Y)$ such that

• $D'$ is upward-closed: if $V \in D'$ and $V \subseteq V'$ is an inclusion of open sets of $Y$, then $V' \in D'$.
• If the union of all the $U_a$ belongs to $D'$, then some finite union $U_{a_1} \cup \ldots \cup U_{a_n}$ belongs to $D'$.

It is not hard to check that $\mathcal{U}$ is indeed a topology. To check that it is strong, we need to check that the membership relation $\in_Y$ is open in $Y \times \mathcal{O}(Y)_{\mathcal{U}}$. Suppose $(x, W) \in \in_Y$, i.e., suppose $x \in W$. If $x$ does not belong to the union of all the $U_a$, then

$$W \times \{W': W \subseteq W'\}$$

is a $\mathcal{U}$-open neighborhood of $(x, W)$ included in $\in_Y$ (by the lemma above). If on the other hand $x$ belongs to some $U_a$, then

$$(W \cap U_a) \times \{W': W \cap U_a \subseteq W'\}$$

is a $\mathcal{U}$-open neighborhood of $(x, W)$, which (again by the lemma) is included in $\in_Y$.

However, the $D \in \mathcal{T}$ with which we started the argument does not belong to $\mathcal{U}$. Suppose otherwise. We know that the union of the $U_a$ contains $V$ and $V \in D$, so $\bigcup_a U_a$ belongs to $D$ by upward closure. Thus some finite union $U_{a_1} \cup \ldots \cup U_{a_n}$ would belong to $D$. But any member of $D$ contains $U$, hence we have

$$C \subseteq U \subseteq U_{a_1} \cup \ldots \cup U_{a_n}$$

contradicting the fact that there is no finite subcover of $C$.

Therefore the topology $\mathcal{T}' = \mathcal{T} \cap \mathcal{U}$ is strictly smaller than $\mathcal{T}$. The intersection $\mathcal{T}'$ of two strong topologies is also strong, so we are done.

$^\dagger$ The unmotivated trick we used above is closely related to the so-called Scott topology on a poset.

There is another neat example (due to Kathleen Lewis) in 1.7.1 of May-Sigurdsson: $\mathbb{Q}\times -$ does not preserve the quotient $\mathbb{Q}\times\mathbb{N} \to \mathbb{Q} \wedge \mathbb{N}$, and therefore the smash product of ordinary topological spaces is not associative: $\mathbb{Q} \wedge (\mathbb{Q}\wedge \mathbb{N}) \ncong (\mathbb{Q}\wedge \mathbb{Q}) \wedge \mathbb{N}$. I haven’t read through the details of the argument, though.