# 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

1. 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~x_2 if there is a y\in Y so that f(y)=x_1 and g(y)=x_2.

No you don’t, but if you identify X under the smallest equivalence relation that includes that relation you just defined, then you get the coequalizer you want.

Comment by Jeremy Henty | November 26, 2007 | Reply

2. […] a lot about the category of topological spaces and continuous maps between them. In particular we’ve seen that it’s complete and cocomplete — it has all limits and colimits. But we’ve […]

Pingback by Topological Groups « The Unapologetic Mathematician | November 27, 2007 | Reply

3. Jeremy, I thought I’d discussed completing a relation to an equivalence relation, hadn’t I? Okay, maybe I need to go back and cover that.

Comment by John Armstrong | November 27, 2007 | Reply

4. John, sorry if my comment was overly blunt. I read “… we get an equivalence relation on X by defining x_1~x_2 if P” as saying that x_1~x_2 is equivalent by definition to P . You may reasonably declare that I misread you since you wrote “if”, not “if and only if” but I think it would be harder to misread if you made it explicit you were completing a relation to an equivalence relation. After all, you were quite explicit whenever you completed a subbase to a topology so I was expecting you to be similarly explicit when constructing the relation. I concede that it’s a judgement call, maybe I was too nitpicky.

Comment by Jeremy Henty | November 28, 2007 | Reply

5. On a separate point, you have a slight LaTeX glitch. You need to write \~ to get a ~ because LaTeX interprets ~ as “unbreakable whitespace”.

Comment by Jeremy Henty | November 28, 2007 | Reply

6. […] sets are open subsets of , but they may not be open as subsets of . But by the definition of the subspace topology, each one must be the intersection of with an open subset of . Let’s just say that each is […]

Pingback by Some compact subspaces « The Unapologetic Mathematician | January 15, 2008 | Reply

7. […] One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family of compact spaces is again compact. Unsurprisingly, the really tough bit comes in […]

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8. Dear sir:
I am student,could you help me please, I need some examples about continuous functions between {regular spaces,normal spaces,competely normal,conncted spaces}i.e let x={a,b,c} and Y={d,e,f,g} such that f-1(f(a))= a,f-1(f(b))={b},f-1(f(c))={c},but .
f-1(f(a,b))=/={a,b},I mean some thing like this.
And I want the proof the invrese image of compact set is compact.
regards

Comment by Gazala | July 2, 2008 | Reply

9. I’m really not sure what you’re going for. Where’s the topology in your example? And what does it have to do with any of those classes of spaces?

As for the inverse image of a compact set being compact, it’s clearly false. Just take a constant function from the reals to the reals. The single point is compact, but its inverse image is the whole line, which isn’t compact.

Comment by John Armstrong | July 2, 2008 | Reply

10. […] is obtained by taking the usual topology on , taking its product with itself, and then taking the quotient topology) by is the divisor of […]

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11. dear sir
if :X—->Y is continuous mapping, if A is a subspace of Y then f-1(A) normal in X
regards

Comment by Gazala | August 4, 2008 | Reply

12. Gazala: again, your setup is clearly false. I’m sure wherever you got the problem from says a lot more, but you don’t seem to understand enough to know what’s part of the statement and what’s not. Either way, I’m not here to do your homework.

Comment by John Armstrong | August 4, 2008 | Reply

13. Though ubiquitous, it’s still a remarkable phenomenon: isn’t it odd how so many flailing students apparently assume the Internet is just bursting with people keen to do others’ tedious homework?

Comment by Sridhar Ramesh | August 4, 2008 | Reply

14. Well, I’m willing to help, but you at least have to pose a problem that makes some sort of sense. As it stands, this is not a problem that any professor would have posed.

Comment by John Armstrong | August 4, 2008 | Reply

15. Of course. What stands out to me is that, often as not, what’s being asked for seems to transcend “Could someone help me understand this?” into “Could someone do this for me?”. But then, I guess that’s always the line one has to watch when giving help, whether asked for or not.

Comment by Sridhar Ramesh | August 4, 2008 | Reply

16. […] know about products of topological spaces. We can take products of metric spaces, too, and one method comes down to us all the way from […]

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17. […] one circle with a marked point for each natural number and quotient by an equivalence relation declaring that all those marked points are really the same point. And […]

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18. […] line and “wrapping” it around itself periodically. I haven’t really mentioned the topologies, but the first approach inherits the subspace topology from the topology on the complex numbers, […]

Pingback by The Circle Group « The Unapologetic Mathematician | May 27, 2009 | Reply

19. […] metric spaces is that we get the same topology as if we’d forgotten the metric and taken the product of topological spaces. This will actually be useful to us, in a way, so I’d like to explain it […]

Pingback by The Topology of Higher-Dimensional Real Spaces « The Unapologetic Mathematician | September 15, 2009 | Reply

20. […] of vectors so that , and for each such define , so . As a slice of the open set in the product topology on , the set is open in . Further, is continuously differentiable on since is continuously […]

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21. […] such a shape in -dimensional space is the product of closed […]

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22. […] space in a natural way, and if this constitutes a subobject in the category. Unfortunately, unlike we saw with topological spaces, it’s not always possible to do this with measurable spaces. But […]

Pingback by Measurable Subspaces I « The Unapologetic Mathematician | April 28, 2010 | Reply

23. […] Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces Before we move on, we want to define some structures that blend algebraic and topological notions. These are all based on vector spaces. And, particularly, we care about infinite-dimensional vector spaces. Finite-dimensional vector spaces are actually pretty simple, topologically. For pretty much all purposes you have a topology on your base field , and the vector space (which is isomorphic to for some ) will get the product topology. […]

Pingback by Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces « The Unapologetic Mathematician | May 12, 2010 | Reply

24. […] – and -dimensional smooth manifolds, respectively, then we can come up with an atlas that makes the product space into an -dimensional smooth manifold, and that it satisfies the conditions to be a product object […]

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25. […] we define an “embedding” to be an immersion where the image — endowed with the subspace topology — is homeomorphic to itself by . This is closer to the geometrically intuitive notion of a […]

Pingback by Immersions and Embeddings « The Unapologetic Mathematician | April 18, 2011 | Reply