The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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