16
$\begingroup$

Is it right to consider assigning a fundamental group to a topological space the same as having a functor from $\mathbf{Top}$ to $\mathbf{Grp}$ ? Are there any other examples of such functors ?

2 Answers 2

29

Assigning the fundamental group to a topological space is definitely a functor. But you have to keep in mind that a fundamental group is always taken with respect to a base point, and hence the functor assigns a pair $(X,x_0)$ consisting of a topological space $X$ and a point $x_0\in X$ to its fundamental group $\pi_1(X,x_0)$. As such, the functor goes from $\mathbf{Top}_\ast$ to $\mathbf{Grp}$.

In more detail, the fundamental group $\pi_1(X,x_0)$ is the group of homotopy classes of loops starting and ending in the base point $x_0$. It is not so hard to show that the map $[\gamma]\mapsto[f\circ\gamma]$ is well-defined for each loop $\gamma$ from $x_0$ to $x_0$ and each morphism $f:(X,x_0)\to(Y,y_0)$ of the category $\mathbf{Top}_\ast$; we take this map to be $\pi_1(f)$. Roughly, this is a definition by post-composition, so it is immediate that this functor respects identity morphisms and compositions.


This is a side remark, because you have been asking about the fundamental group explicitly. But I feel it is in place here because it is natural to consider a functor with domain $\mathbf{Top}$ that is `like taking the fundamental group'.

Instead of $\mathbf{Top}_\ast\to\mathbf{Grp}$, one could also work with a functor $\mathbf{Top}\to\mathbf{Grpd}$, where the category $\mathbf{Grpd}$ is the category of groupoids (categories in which all morphisms are isomorphisms). The functor sends a topological space $X$ to the groupoid which has the points of $X$ as objects and between two points $x$ and $y$ of $X$ the morphisms are the homotopy classes of paths from $x$ to $y$. This gives you the fundamental groupoid rather than the fundamental group.

The n-lab has more information on the fundamental groupoid.


There are many more examples of functors from $\mathbf{Top}$ or related categories. An important one is the singular functor to the category $\mathbf{Sset}$ of simplicial sets. The category if simplicial sets is defined as follows: first you consider the category $\Delta$ consisting of an object $[n]$ for each natural number $n$, where $[n]$ is the partial ordered set $\{0,\ldots,n\}$ with the usual order; the morphisms are the order preserving maps. Then $\mathbf{Sset}$ is the category of contravariant functors from $\Delta$ to $\mathbf{Set}$.

For each natural number $n$, there is the topological space $$ |\Delta^n|:=\big\{(t_0,\ldots,t_n)\in[0,1]^{n+1}:\textstyle\sum_{i=0}^n t_i=1\big\}, $$ which is called the standard $n$-simplex. To test your understanding of these definitions, you can show that the map $[n]\mapsto|\Delta^n|$ is a functor from $\Delta$ to $\mathbf{Top}$. Now we can define the functor $S:\mathbf{Top}\to\mathbf{Sset}$, which is called the singular functor, by assigning to each topological space $X$ the functor $$ n\mapsto\mathbf{Top}(|\Delta^n|,X) $$ It turns out that the simplicial sets $S(X)$ have very nice properties. One of them is that they really are $\infty$-groupoids. Also, the set $\mathbf{Top}(|\Delta^n|,X)$ is used to define the $n$-th homology group of $X$, which is gives yet another functor from the category of topological spaces. All of these functors have been (and are) important for the investigation of topological spaces.

  • 0
    Thanks, this is very nice... I cannot upvote apparently but I would if I could2012-06-03
  • 0
    @JeanThiviers: it's a bit strange that you can't upvote it, but don't worry about it :) I included a link to the n-lab page where you can learn more about fundamental groupoids.2012-06-03
  • 2
    Lovely answer, thank you. By the way, this reminds me of a question that came up recently, but I couldn't find the answer online: is the fundamental group functor continuous and cocontinuous? I mean, I think I know it's cocontinuous, that's given by the Seifert-van Kampen theorem. And I know it preserves products. But what about general inverse limits?2012-06-04
13

Sort of. Technically, a fundamental group is assigned to a pointed topological space, that is a topological space with a distinguished point (the basepoint of your loops). The category of such spaces is denoted $\textbf{Top}_\ast$, and to make a functor from there to $\textbf{Grp}$, you not only need to assign a group to each space (i.e. the fundamental group) but also assign to each morphism in $\textbf{Top}_\ast$ a morphism in $\textbf{Grp}$, in a way that is compatible with composition of morphisms and sends the identity to the identity.

In this context, this means assigning to each continuous map $f$ that preserves basepoints (i.e. maps the basepoint of one space to the basepoint of the other) a homomorphism $f_*$ of the corresponding fundamental groups, such that $\operatorname{id}_* = \operatorname{id}$ and $(f\circ g)_* = f_* \circ g_*$. We can do this by saying, if $[u]$ is the homotopy class of the path $[u]$, then $f_*[u] = [f\circ u]$. Hence we can define a functor using the fundamental group.

The homology groups are also functors, this time from triangulable topological spaces (not pointed!) to abelian groups (modulo some complications; see the comments).

  • 3
    Technically, you need to work in the category of *pointed* spaces.2012-06-03
  • 0
    @SteveD oh right, yes. My category theory knowledge is a bit scattered – is there a standard notation for that category? (feel free to edit my answer if you think you can make it more correct)2012-06-03
  • 0
    @benmachine I edited your answer to clarify that one needs to work in the category of pointed topological spaces; I hope this is OK.2012-06-03
  • 4
    Standard notation is $\textbf{Top}_*$, actually.2012-06-03
  • 0
    @ZhenLin Yes, I wanted to write that but I forgot the LaTeX code for it. The closest symbol to that for which I knew the LaTeX code was $\bigodot$ which is why I used $\bigodot$ instead.2012-06-03
  • 0
    @AmiteshDatta: you can look at other people's LaTeX code by right clicking on the math and then `Show Math As > TeX commands`2012-06-03
  • 0
    @AmiteshDatta: yes, that's fine, thank you :)2012-06-03
  • 0
    @SteveD, just out of interest, is there any difference between `\ast` and a plain `*`? (I went ahead and rewrote large parts of the answer anyway, because the way I first wrote it was somewhat lazily structured)2012-06-03
  • 0
    @benmachine: `*` has a marginally larger likelihood of being eaten by a grue (i.e. Markdown).2012-06-03
  • 0
    Singular homology doesn't require a triangulation/CW complex/whatever, does it?2012-06-03
  • 0
    @DylanMoreland I've heard that, but I've only ever learnt simplicial homology :P2012-06-03
  • 0
    No singular homology works for any top. space.2012-06-03
  • 0
    You need to be a little careful when discussing *simplicial homology* as a functor from the category of triangulable topological spaces to the category of abelian groups. The reason is that a triangulable space may admit many triangulations and although the simplicial groups depend only on the isomorphism class of the triangulable space, the simplicial groups corresponding to different triangulations may very well be different (isomorphic) groups. For example, when you wish to define the homomorphism induced by a continuous map, you need to *choose* the triangulations.2012-06-04
  • 0
    The way to avoid this problem is standard. If $X$ is a topological space, then let $S$ be the "set" of all triangulations of $X$. (Of course, this is not a set but rather a class; however, every simplicial complex triangulating $X$ is isomorphic to a simplicial complex imbedded in $E^{\left|X\right|}$ where $E^{\left|X\right|}$ is the space of tuples of length the cardinality $\left|X\right|$ of $X$ with all but finitely many terms equal to $0$. Of course, $E^{\left|X\right|}$ is a set.) If $S=(K_{\alpha},h_{\alpha})_{\alpha\in A}$, then consider the disjoint union of $H_p(K_{\alpha})$ ...2012-06-04
  • 0
    ... We can define an equivalence relation on this disjoint union by setting $x_{\alpha}\equiv y_{\beta}$ if and only if $(h_{\beta}^{-1}\circ h_{\alpha})_{\ast}(x_{\alpha})=y_{\beta}$. The resulting set, which we **define to be $H_p(X)$**, has the unique structure of a group such that the canonical map $H_{p}(K_{\alpha})\to H_p(X)$ is an isomorphism of groups for all $\alpha\in A$. You can then check that this definition of $H_p(X)$ is functorial with respect to continuous maps and we define these groups to be the **simplicial homology groups of the triangulable space $X$**.2012-06-04
  • 0
    @AmiteshDatta: I didn't really feel up to putting all that in the answer, so I just referred to your comment :P2012-06-04