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A very basic definition in category theory is the definition of morphism between objects. If the category is a construct, i.e., a category $\mathcal C$ equipped with a faithful functor $U\colon \mathcal C\to Set$, how can the morphisms be seen as structure-preserving?

In other words, what exactly does "structure-preserving" mean? If I have two topological spaces, why are continuous functions "structure-preserving" functions?

Has it something to do with the "final topology" or "initial topology" of a continuous function?

User
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    Your definition of morphism is incorrect. In some categories, morphisms are not even functions. – Ink Mar 23 '14 at 02:45

3 Answers3

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The definition of a category does not talk about any structure of the objects. All you need is a class of objects, a class of morphisms and some rule to compose morphisms.

It just happens that taking topological spaces as objects, continuous maps as morphisms and the ordinary composition of maps as the composition of morphisms turns out to be a category, namely $\mathbf{Top}$. In a category, two objects are isomorphic if there are mutually inverse morphisms between them. In the context of topological spaces and continuous maps this turns out to describe homeomorphic spaces, thus "isomorphism" translates to "homeomorphism" in this case. So any statement that is true in category theory "up to isomorphism" will be true in this category "up to homeomorphism".

You could as well form a category by again taking the topological spaces as objects but instead of continuous maps take all maps as morphisms. Isomorphisms in this category will just be bijective maps, so $S^1$ and $\mathbb R$ will be isomorphic objects in this category, but not in the category of topological spaces and continuous maps. This category is really just the category of sets, namely $\mathbf{Set}$, since the morphisms don't capture any topological features of the spaces.

For most structures in mathematics, we have some idea of two of a kind being "isomorphic", for topological spaces that is homeomorphic spaces, for groups it's isomorphic groups, for vector spaces it's isomorphic vector spaces. To capture these "structural properties" in a category you need to pick the class of morphism so your definition of "isomorphism" becomes "$A$ and $B$ are isomorphic if there are mutually inverse morphisms between them". For topological spaces you need to choose continuous maps to capture homeomorphism of spaces as isomorphism of objects.

Another example is the homotopy category of topological spaces $\mathbf{hTop}$: If we want two topological spaces to be "isomorphic" in our category whenever they are homotopic as topological spaces, we need a different class of morphisms. This time what we need is equivalence classes of homotopic continuous maps, with a well defined composition of these equivalence classes. Then we have "$A$ and $B$ are isomorphic objects" if and only if $A$ and $B$ are homotopic spaces, so this category captures the homotopy class of spaces.

Christoph
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    This does not really answer my question...My question is : what makes a continuous function a morphism in this category , and what makes a non continuous function not a morphism? What does it mean for a function to be a morphism(i.e., what does it mean preserving the structure?) – User Mar 22 '14 at 19:13
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    The continuous maps are the morphisms in $\mathbf{Top}$ by definition. Why that definition is useful is explained in my answer. It captures homeomorphism of spaces as isomorphism of objects. – Christoph Mar 22 '14 at 19:17
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    @user132951 The above answer does not give a good motivation of why we should "pick" certain morphisms to be the structure preserving maps. Every small category $\mathcal C$ in fact has a construct structure where the underlying functor is given by $U\colon \mathcal C\to \mathbf{Set}$, $C\mapsto {f\in Ar(\mathcal C): cod(f)=C}$ and arrows by post-composition. We need to be able to point to what the structure on each set is. Then try to relate these structures so that there is some preservation of structure. This answer fails to explain that. Please consider. – Rachmaninoff Mar 23 '14 at 03:40
  • @Christoph In what sense is $\mathbf{hTop}$ a category consisting of structured-sets and structure preserving maps? The category $\mathbf{hTop}$ is defined by an equivalence on hom-sets. How do you define the forgetful functor on equivalence classes of hom-sets? Is there a canonical choice such that commutativity conditions are satisfied? What structure is being preserved in the case of $\mathbf{hTop}$? I believe this is a poor example of structured sets. If the question was about simply a distinction between arrows and set functions (often misunderstood), I could agree with your answer. – Rachmaninoff Mar 23 '14 at 04:53
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    @Rachmaninoff The OP is most probably just starting to learn category theory and needs some motivation on how to think about morphisms and why different classes of morphisms give different categories and which one might be the "right" choice in the case of topological spaces. That's what I elaborated on. In your answer your motivation is that $f^{-1}:\tau'\to\tau$ is a bounded poset map, how is that a better motivation for a beginner in cat theory? – Christoph Mar 23 '14 at 07:28
  • @Rachmaninoff I didn't say anything about forgetful functors and didn't pretend there was a canonical choice one on $\mathbf{hTop}$, I just pointed out a different class of morphisms on the same class of objects might give us a different category that now captures homotopy instead of homeomorphy. – Christoph Mar 23 '14 at 07:29
  • @Christoph I am attempting to answer the question which is about "structured sets" and not "category" per se. I believe I gave appropriate examples about what a structure is on a set, both in terms of algebraic and topological. I also gave links to where the OP may refer to. If the OP knows any topology, (s)he should know that $f^{-1}$ takes open sets to open sets. It is a small leap to show that $f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)$ and for unions as well. The use of the term "construct" by the OP suggests some knowledge of the concept. – Rachmaninoff Mar 23 '14 at 07:36
  • @Christoph Then $\mathbf{hTop}$ is not an example of a construct. See http://tac.mta.ca/tac/reprints/articles/6/tr6abs.html – Rachmaninoff Mar 23 '14 at 07:38
  • @Christoph in other word, there is no subclass of set functions which describe the arrows in $\mathbf{hTop}$ – Rachmaninoff Mar 23 '14 at 08:01
  • In other words, the objects of $\mathbf{hTop}$ are NOT structured sets with structure preserving morphisms. – Rachmaninoff Mar 23 '14 at 09:03
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I think of a structured set as a set $X$ along with some other information telling me how $X$ is "structured". For example, a group is a set $G$ along with its "structure" which is a binary operation $+\colon G\times G \to G$, a unary operation $-\colon G \to G$ and a nullary operation $e\colon 1\to G$ that satisfy the group axioms. Then a group structure-preserving map $h\colon \langle G;+,-,e\rangle \to \langle G';+',-',e'\rangle$ is a set map $h\colon G\to G'$ that preserves the structure, i.e., $e'=he$, $+'\circ(h\times h)=h\circ+$, and $-'\circ h=h\circ -$. To study how this structure behaves on its underlying set, we look at the forgetful functor. In the case of groups, $U\colon \langle G;+,-,e\rangle \mapsto G$, $h\mapsto h$.

So in the case of topological spaces, we have different ways of describing the structure on its underlying set, be it by open sets, closed sets, the Kuratowski closure operator, or neighborhood description (see Peter Clark's notes). These formally give us different categories, the objects being sets equipped with one of these descriptions as its structure and the arrows being those set maps which respect the structure (more on this later). They all are equipped with a forgetful functor, and moreover they are concretely isomorphic, meaning the there are isomorphic functors which respect the forgetful functors.

For example, if $\mathbf{Top}_{open}$ is the category of topological spaces defined by open sets and $\mathbf{Top}_{closed}$ is the category of topological spaces defined by closed sets, there is an obvious isomorphism $F\colon \mathbf{Top}_{open} \to \mathbf{Top}_{closed}$, $\langle X; \tau\rangle\mapsto \langle X; \overline{\tau}\rangle$ where the set of open sets $\tau$ map to their set complements in $\overline{\tau}$. Then the forgetful functors $U\colon \mathbf{Top}_{open}\to \mathbf{Set}$ and $U'\colon \mathbf{Top}_{closed}\to \mathbf{Set}$ are related by this iso, i.e., $U=U'F$. We say they are concretely isomorphic meaning the structure is essentially the same.

As to your question about why the topological maps are said to be structure preserving, it is because the maps respect the structure of the topological spaces. If we take the open set definition of topological spaces, then $f\colon \langle X; \tau\rangle \to \langle X';\tau' \rangle$ is a set map $f\colon X\to X'$ such that there is a preservation of the open set structure, ie, $f^{-1}\colon \tau'\to \tau$ is a bounded poset map.

It turns out that there are topological-like constructs and algebraic-like constructs, depending on what the fibers of the forgetful functor looks like on objects.

For example, let's look at the the fiber of $U\colon \mathbf{Top}\to \mathbf{Set}$ over a set $X$. We can define a poset structure on the fiber by setting $\langle X;\tau\rangle \leq \langle X;\tau'\rangle$ iff the underlying identity map is a continuous map $id_X\colon \langle X;\tau\rangle \to \langle X;\tau'\rangle$. This is where your final and cofinal topologies come in. We have every fiber equipped with a bounded poset structure with the bottom being the cofinal topology on that set and top being the final topology. This is because the identity from a set with the cofinal topology (discrete topology) is continuous and the identity to a set with the final topology (indiscrete topology) is continuous.

In general, topological-like structures have nondiscrete posets as fibers and algebraic-like structure have discrete posets as fibers where two structures $\langle X;S\rangle \leq \langle X;S'\rangle$ iff the underlying identity map is a structure preserving map $id_X\colon \langle X;S\rangle \to \langle X;S'\rangle$. See The Joy of Cats for a more detailed discussion http://katmat.math.uni-bremen.de/acc/acc.pdf

Martin Sleziak
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Rachmaninoff
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  • Thanks for the time answering my question first. I am aware that continuous functions preserve opens in the topology when taking the inverse. Even though continuous functions are structure preserving in \mathbf{Top}, what it really seems to me is that the inverse of a continuous function is structure preserving.

    To me, this sounds very asymmetrical to what happens in \mathbf{Grp}

    \mathbf{Top} morphisms "preserve" the structure of the topology on inverse image while morphisms in \mathbf{Grp} preserve the structure through the image.

     – User Mar 23 '14 at 09:51
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    @user132951 That is a very good observation. A good rule of thumb is that categories of geometry (like topology) are somehow dual to that of categories of algebra. But maybe a better explanation is that vicinity of neighborhoods of points are preserved. Recall that a real function $f\colon U\to \mathbb R$ is continuous iff $\lim_{x_0\to x}f(x_0)=f(x)$ for each $x\in X$. We can generalize a bit by looking at filters. See http://en.wikipedia.org/wiki/Filter_(mathematics)#Functions_on_topological_spaces – Rachmaninoff Mar 23 '14 at 10:17
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    Once we have the notion of filter bases down, then a function $f\colon X\to Y$ is continuous iff the image of the filter base on $X$ forms a filter base on $Y$. – Rachmaninoff Mar 23 '14 at 10:24
  • But of course, this is equivalent to the definition given by the inverse of an open set is open. There are of course others. I just wanted to give you one where the structure preservation was going in the same direction as the set function. – Rachmaninoff Mar 23 '14 at 10:27
  • Are you telling me that they also "preserve" structure though image because of filters? I have some backgrounds concerning filters so if there is a full explanation of this I can grab it.

    Or maybe I have misunderstood.

    My final though is : By counterimage topologies are preserved by though open sets and by image thanks to the filters. is it is it just the first part

     – User Mar 23 '14 at 10:39
  • I should have added that this image of the filter base clusters at $f(x)$. I think the wiki article explains it well. Else look at Willard's text General Topology – Rachmaninoff Mar 23 '14 at 11:12
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Good question.

In algebra, "structure-preserving" is easy to define, because algebraic structures can be viewed as product-preserving functors out of Lawvere theories, and the homomorphisms between them are just the natural transformations.

However outside of algebra, it becomes much trickier to work out what is the "correct" notion of homomorphism, and sometimes there is more than one "correct" answer, depending on what you're trying to do.

goblin GONE
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  • What is tricky about the notion of preserving neighborhoods of points? So a set map $f\colon X\to Y$ must respect "nearness", i.e., open neighborhoods of $f(x)$ in $Y$ must have come from open sets in $X$. – Rachmaninoff Mar 23 '14 at 05:09
  • @Rachmaninoff well why not reflect neighbourhoods? Why not preserve neighbourhoods up to expansion? etc. – goblin GONE Mar 23 '14 at 05:20
  • That's right. But if I understand what you mean by reflect neighbourhoods, or expansion, all of these definitions are concretely isomorphic. So they are all essentially describing the same structure-preservation, which is what I mentioned in the answer above. – Rachmaninoff Mar 23 '14 at 05:25
  • If "reflecting neighbourhoods" and "preserving neighbourhoods up to expansion" give us different morphisms, then we should be able to point to the structure in the definition of a structured set...otherwise these are not structure-preserving... – Rachmaninoff Mar 23 '14 at 05:40