In the category of topological spaces ($\mathbf{Top}$), products do not always preserve colimits. If they did then $\mathrm{Hom}_\mathbf{Top}(-\times X,S)$ would be representable and hence $\mathbf{Top}$ would be Cartesian closed (which it isn't). I think that products do preserve coproducts, so it must be that there's some coequaliser which products don't preserve. I'm trying to understand why this is in more concrete terms, but I've struggled to find a simple example that I can examine in detail.

What are some simple spaces $A$, $B$ and $X$ and maps $f,g:A\to B$ in $\mathbf{Top}$ such that the product of $X$ with the coequaliser is different from the coequaliser of the products?

The same question for the category of locales is here.

  • $\begingroup$ A different angle of attack to the underlying question of why Top is not cartesian closed: if it were, then the underlying set of $Y^X$ would have to be $Hom_{Top}(1, Y^X) \simeq Hom_{Top}(X, Y)$, the set of continuous functions. And given a net $f : I \to Y^X$, $f_i \to f_0$ if and only if the map $I \cup \{ \infty \} \to Y^X$ is continuous, if and only if the map $(I \cup \{ \infty \}) \times X \to Y$, which would allow us to define what the closure operator on $Y^X$ would have to be. Then, come up with an example where this isn't idempotent. $\endgroup$ Oct 24, 2018 at 18:14
  • $\begingroup$ @DanielSchepler Thanks, but I'm actually interested in the case of coequalisers in particular. I just mentioned Cartesian closedness to explain how I knew there must be an example of one that doesn't commute with products. $\endgroup$ Oct 24, 2018 at 19:10
  • 3
    $\begingroup$ Hmm, according to ncatlab.org/nlab/show/exponential+law+for+spaces it must be that if $X$ is not a core-compact space, for example $X = \mathbb{Q}$, then the functor $X \times - : Top \to Top$ does not preserve coequalizers. That doesn't really give the full proof but it might at least be somewhere to start. I think I also saw somewhere that the Sierpinski space $(\{ 0, 1 \}, \{ \emptyset, \{ 1 \}, \{ 0, 1 \} \})$ might be involved in a counterexample. $\endgroup$ Oct 24, 2018 at 21:44

1 Answer 1


(Adapted from Ronald Brown's 'Topology and Groupoids', Section 4.3 Example 4, Page 111.)

Consider $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$ with their usual topologies. Let $i:\mathbb{Z}\hookrightarrow\mathbb{R}$ be the usual inclusion, and define $j :\mathbb{Z}\to\mathbb{R}$ by $j(n) = i(n+1)$. Our example of failed preservation will be that the canonical map $$\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})\to\mathrm{coeq}(i,j)\times\mathbb{Q}$$ is not a homeomorphism.

Since the forgetful functor $\mathbf{Top}\to\mathbf{Set}$ preserves both limits and colimits, the underlying function of this map is indeed a bijection. The underlying set of both spaces is the quotient of $\mathbb{R}\times\mathbb{Q}$ by the equivalence relation that relates $(r,q)$ to $(r',q)$ whenever both $r$ and $r'$ are integers. The reason this bijection is not a homeomorphism is that there are open sets in $\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})$ whose images are not open in $\mathrm{coeq}(i,j)\times\mathbb{Q}$.

To construct such an open set, consider the graphs of two continuous functions $f,g:\mathbb{R}\to\mathbb{R}$ with the following properties:

  • Both $f(x)$ and $g(x)$ are strictly positive for all $x$, but tend to $0$ as $x$ tends to $+\infty$ and $-\infty$.

  • We have $f(x)=g(x)$ iff $x$ is an integer, and in this case $f(x)$ and $g(x)$ are irrational.

For example we could take $f(x)=\frac{\pi+\sin(\pi x)}{1+x^2}$ and $g(x)=\frac{\pi-\sin(\pi x)}{1+x^2}$. Now define $U$ to be the image under the quotient map of the subset of $\mathbb{R}\times\mathbb{Q}$ containing the points $(r,q)$ for which $q$ is either less than both $f(r)$ and $g(r)$ or greater than both $f(r)$ and $g(r)$.

Graph of counterexample

Then $U$ is open in $\mathrm{coeq}(i\times\mathbb{Q},j\times\mathbb{Q})$ since its preimage under the quotient map is open in $\mathbb{R}\times\mathbb{Q}$. But it is not open in $\mathrm{coeq}(i,j)\times\mathbb{Q}$ since every neighbourhood of $0$ in $\mathrm{coeq}(i,j)$ contains arbitrarily large nonintegers and hence every open rectangle around $(0,0)$ in $\mathrm{coeq}(i,j)\times\mathbb{Q}$ meets the area between the graphs of $f$ and $g$.


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