# The Unapologetic Mathematician

## Limits of Topological Spaces

We’ve defined topological spaces and continuous maps between them. Together these give us a category $\mathbf{Top}$. We’d like to understand a few of our favorite categorical constructions as they work in this context.

First off, the empty set $\varnothing$ has a unique topology, since it only has the one subset at all. Given any other space $X$ (we’ll omit explicit mention of its topology) there is a unique function $\varnothing\rightarrow X$, and it is continuous since the preimage of any subset of $X$ is empty. Thus $\varnothing$ is the initial object in $\mathbf{Top}$.

On the other side, any singleton set $\{*\}$ also has a unique topology, since the only subsets are the whole set and the empty set, which must both be open. Given any other space $X$ there is a unique function $X\rightarrow\{*\}$, and it is continuous because the preimage of the empty set is empty and the preimage of the single point is the whole of $X$, both of which are open in $X$. Thus $\{*\}$ is a terminal object in $\mathbf{Top}$.

Now for products. Given a family $X_\alpha$ of topological spaces indexed by $\alpha\in A$, we can form the product set $\prod\limits_{\alpha\in A}X_\alpha$, which comes with projection functions $\pi_\beta:\prod\limits_{\alpha\in A}X_\alpha\rightarrow X_\beta$ and satisfies a universal property. We want to use this same set and these same functions to describe the product in $\mathbf{Top}$, so we must choose our topology on the product set so that these projections will be continuous. Given an open set $U\subseteq X_\beta$, then, its preimage $\pi_\beta^{-1}(U)$ must be open in $\prod\limits_{\alpha\in A}X_\alpha$. Let’s take these preimages to be a subbase and consider the topology they generate.

If $X$ is any other space with a family of continuous maps $f_\alpha:X\rightarrow X_\alpha$, then the universal property in $\mathbf{Set}$ gives us a unique function $f:X\rightarrow\prod\limits_{\alpha\in A}X_\alpha$. But will it be a continuous map? To check this, remember that we only need to verify it on a subbase for the topology on the product space, and we have one ready to work with. Each set in the subbase is the preimage $\pi_\beta^{-1}(U)$ of an open set in some $X_\beta$, and then its preimage under $f$ is $f^{-1}(\pi_\beta^{-1}(U))=(\pi_\beta\circ f)^{-1}(U)=f_\beta^{-1}(U)$, which is open by the assumption that each $f_\beta$ is continuous. And so the product set equipped with the product topology described above is the categorical product of the topological spaces $X_\alpha$.

What about coproducts? Let’s again start with the coproduct in $\mathbf{Set}$, which is the disjoint union $\biguplus\limits_{\alpha\in A}X_\alpha$, and which comes with canonical injections $\iota_\beta:X_\beta\rightarrow\biguplus\limits_{\alpha\in A}X_\alpha$. This time let’s jump right into the universal property, which says that given another space $X$ and functions $f_\alpha:X_\alpha\rightarrow X$, we have a unique function $f:\biguplus\limits_{\alpha\in A}X_\alpha\rightarrow X$. Now we need any function we get like this to be continuous. The preimage of an open set $U\subseteq X$ will be the union of the preimages of each of the $f_\alpha$, sitting inside the disjoint union. By choosing $X$, the $f_\alpha$, and $U$ judiciously, we can get the preimage $f_\alpha^{-1}(U)$ to be any open set we want in $X_\alpha$, so the open sets in the disjoint union should consist precisely of those subsets $V$ whose preimage $\iota_\alpha^{-1}(V)\subseteq X_\alpha$ is open for each $\alpha\in A$. It’s easy to verify that this collection is actually a topology, which then gives us the categorical coproduct in $\mathbf{Top}$.

If we start with a topological space $X$ and take any subset $S\subseteq X$ then we can ask for the coarsest topology on $S$ that makes the inclusion map $i:S\rightarrow X$ continuous, sort of like how we defined the product topology above. The open sets in $S$ will be any set of the form $S\cap U$ for an open subset $U\subseteq X$. Then given another space $Y$, a function $f:Y\rightarrow S$ will be continuous if and only if $i\circ f:Y\rightarrow X$ is continuous. Indeed, the preimage $(i\circ f)^{-1}(U)=f^{-1}(S\cap U)$ clearly shows this equivalence. We call this the subspace topology on $S$.

In particular, if we have two continuous maps $f:X\rightarrow Y$ and $g:X\rightarrow Y$, then we can consider the subspace $E\subseteq X$ consisting of those points $x\in X$ satisfying $f(x)=g(x)$. Given any other space $Z$ and a continuous map $h:Z\rightarrow X$ such that $f\circ h=g\circ h$, clearly $h$ sends all of $Z$ into the set $E$; the function $h$ factors as $e\circ h'$, where $e:E\rightarrow X$ is the inclusion map. Then $h'$ must be continuous because $h$ is, and so the subspace $E$ is the equalizer of the maps $f$ and $g$.

Dually, given a topological space $X$ and an equivalence relation $\sim$ on the underlying set of $X$ we can define the quotient space $X/\sim$ to be the set of equivalence classes of points of $X$. This comes with a canonical function $p:X\rightarrow X/\sim$, which we want to be continuous. Further, we know that if $g:X\rightarrow Y$ is any function for which $x_1\sim x_2$ implies $g(x_1)=g(x_2)$, then $g$ factors as $g=g'\circ p$ for some function $g':X/\sim\rightarrow Y$. We want to define the topology on the quotient set so that $g$ is continuous if and only if $g'$ is. Given an open set $U\in Y$, its preimage $g'^{-1}(U)$ is the set of equivalence classes that get sent into $U$, while its preimage $g^{-1}(U)$ is the set of all points that get sent to $U$. And so we say a subset $V$ of the quotient space $X/\sim$ is open if and only if its preimage — the union of the equivalence classes in $V$ is open in $X$.

In particular, if we have two maps $f:Y\rightarrow X$ and $g:Y\rightarrow X$ we get an equivalence relation on $X$ by defining $x_1\sim x_2$ if there is a $y\in Y$ so that $f(y)=x_1$ and $g(y)=x_2$. If we walk through the above description of the quotient space we find that this construction gives us the coequalizer of $f$ and $g$.

And now, the existence theorem for limits tells us that all limits and colimits exist in $\mathbf{Top}$. That is, the category of topological spaces is both complete and cocomplete.

As a particularly useful example, let’s look at an example of a pushout. If we have two topological spaces $U$ and $V$ and a third space $A$ with maps $A\rightarrow U$ and $A\rightarrow V$ making $A$ into a subspace of both $U$ and $V$, then we can construct the pushout of $U$ and $V$ over $A$. The general rule is to first construct the coproduct of $U$ and $V$, and then pass to an appropriate coequalizer. That is, we take the disjoint union $U\uplus V$ and then identify the points in the copy of $A$ sitting inside $U$ with those in the copy of $A$ sitting inside $V$. That is, we get the union of $U$ and $V$, “glued” along $A$.

November 26, 2007 Posted by | Category theory, Topology | 25 Comments