Simple example of product not preserving coequaliser in $\mathbf{Top}$

Solution 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 $g(r)$.

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 $g$.