What does Structure-Preserving mean?

Solution 1:

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.

Solution 2:

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 Kurotowski 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

Solution 3:

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.