The very first algebraic invariant one learns in topology is probably the fundamental group of a space. But it always bothers me that the construction of the fundamental group requires the choice of a base point. Yes, of course there is an isomorphism \(\pi _ 1(X, x _ 0) \simeq \pi _ 1(X, x _ 1)\) for any two \(x _ 0, x _ 1 \in X\) given that \(X\) is path connected. But this isomorphism is not canonical; it again involves the choice of a path \(\gamma : [0,1] \rightarrow X\) which starts at \(x _ 0\) and ends at \(x _ 1\), and different paths \(\gamma\) result in different isomorphisms \(\varphi _ \gamma : \pi _ 1(X, x _ 0) \rightarrow \pi _ 1(X, x _ 1)\). So it can be safely said that the typical definition of the fundamental group depends highly on the base point. As Grothendieck put it in his Esquisse d’un Programme,
… people still obstinately persist, when calculating with fundamental groups, in fixing a single base point, instead of cleverly choosing a whole packet of points which is invariant under the symmetries of the situation, which thus get lost on the way.
The exposition of this post will roughly follow the order in which covering spaces are presented in the classic Mun00. The wordings of many theorems are borrowed from that book.
1. Basic definitions#
Let us start by defining a groupoid.
Definition 1. A groupoid \(\mathcal{G}\) is an algebraic structure consisting of:
- a set \(\mathrm{ob}(\mathcal{G})\) of objects,
- a set \(\hom(x,y)\) of morphisms (also called arrows) for any two objects \(x, y \in \mathrm{ob}(\mathcal{G})\),
- for each \(x \in \mathrm{ob}(\mathcal{G})\), a designated morphism \(\mathrm{id} _ x \in \hom(x,x)\),
- for each \(x, y, z \in \mathrm{ob}(\mathcal{G})\), a composition map \[ \displaystyle \mathrm{comp} _ {x,y,z} : \hom(y,z) \times \hom(x,y) \rightarrow \hom(x,z),\]
- for each \(x, y \in \mathrm{ob}(\mathcal{G})\), a inverse map \[ \displaystyle \mathrm{inv} _ {x,y} : \hom(x,y) \rightarrow \hom(y,x)\]
satisfying, for any \(f \in \hom(x,y)\), \(g \in \hom(y,z)\), and \(h \in \hom(z,w)\) the properties:
- \(f \cdot \mathrm{id} _ x = \mathrm{id} _ y \cdot f = f\),
- \(h \cdot (g \cdot f) = (h \cdot g) \cdot f\),
- \(f^{-1} \cdot f = \mathrm{id} _ x\) and \(f \cdot f^{-1} = \mathrm{id} _ y\),
where for notational convenience we denote \(g \cdot f = \mathrm{comp} _ {x,y,z}(g, f)\) and \(f^{-1} = \mathrm{inv} _ {x,y}(f)\).
Although the definition looks awfully convoluted, one can intuitively think a groupoid as a group-like structure; the composition of morphisms associate, and they all have inverses, but it might not be able to compose two morphisms. Anyone familiar with category theory will immediately notice that a groupoid is merely a category in which every morphism is an isomorphism. We will denote \(f : x \rightarrow y\) to mean \(f \in \hom(x,y)\). So for instance, a group is simply a groupoid with a single object. In fact, for any element \(x \in \mathrm{ob}(\mathcal{G})\), the set \(\hom(x, x)\) is always a group. This follows from the fact that multiplication is associative and there always is an inverse. Also, if \(\hom(x,y)\) is nonempty, then \(\hom(x,x)\) and \(\hom(y,y)\) is isomorphic. Indeed, if \(f \in \hom(x,y)\) then the map \(g \mapsto f g f^{-1}\) gives an isomorphism \(\hom(x,x) \rightarrow \hom(y,y)\).
Definition 2. Consider a topological space \(X\). The fundamental groupoid \(\Pi _ 1(X)\) is defined as the following:
- the objects \(\mathrm{ob}(\Pi _ 1(X))\) consists of points of \(X\),
- the arrows \(\hom(x,y)\) consists of all homotopy classes of paths \(\gamma : [0,1] \rightarrow X\) such that \(\gamma(0) = x\) and \(\gamma(1) = y\).
Composition and inversion are defined in the most natural way; for two homotopy classes \([f] : x \rightarrow y\) and \([g] : y \rightarrow z\) the composition will be just \([g] \ast [f]\), and the inverse will be \([f^{-1}]\). One huge benefit of considering the fundamental groupoid is that we do not have to specify a base point.
For any \(x _ 0 \in X\), it is clear that \(\hom(x _ 0,x _ 0) = \pi _ 1(X, x _ 0)\). This means that the fundamental groupoid contains all information about the fundamental group, as well as the space itself. It turns out that the fundamental groupoid is extremely useful when dealing with covering spaces.
Definition 3. A groupoid \(\mathcal{G}\) is called path connected if for any \(x, y \in \mathcal{G}\) the set \(\hom(x,y)\) is always nonempty. It is an easy exercise to show that \(X\) is path connected if and only if the fundamental groupoid \(\Pi _ 1(X)\) is path connected.
One can also define morphisms between groupoids. In the category theoretic interpretation, these are just functors between groupoids.
Definition 4. Consider two groupoids \(\mathcal{G}\) and \(\mathcal{H}\). A groupoid morphism \(F : \mathcal{G} \rightarrow \mathcal{H}\) consists of:
- a map \(F : \mathrm{ob}(\mathcal{G}) \rightarrow \mathrm{ob}(\mathcal{H})\),
- for each \(x, y \in \mathrm{ob}(\mathcal{G})\), a map \(F : \hom(x,y) \rightarrow \hom(F(x), F(y))\),
which satisfies \(F(\mathrm{id} _ x) = \mathrm{id} _ {F(x)}\) for each \(x \in \mathrm{ob}(\mathcal{G})\), and \(F(g \cdot f) = F(g) \cdot F(f)\) for each \(f : x \rightarrow y\) and \(g : y \rightarrow z\). Note that it is possible to compose two groupoid morphisms.
2. Coverings of groupoids#
From now on, we shall consider only path connected spaces and groupoids unless noted otherwise. Let \(X\) and \(Y\) be two spaces. Suppose there is a continuous map \(p : X \rightarrow Y\). Just as \(p\) induces a canonical homomorphism \(p _ \ast : \pi _ 1(X, x _ 0) \rightarrow \pi _ 1(Y, y _ 0)\), so it induces a groupoid morphism \(p _ \ast : \Pi _ 1(X) \rightarrow \Pi _ 1(Y)\). In fact, the fundamental groupoid is a functor from the category \(\mathsf{Top}\) of topological spaces to \(\mathsf{Grpd}\) of groupoids. In particular, let us consider a covering map \(p : E \rightarrow B\). One important—probably, the most important—property of covering maps is the unique lifting property.
Theorem 5 (Unique lifting). Let \(p : E \rightarrow B\) be a covering map, and let \(p _ \ast : \Pi _ 1(E) \rightarrow \Pi _ 1(B)\) be the canonical groupoid morphism. Then for each pair of morphism \(f : b _ 0 \rightarrow b _ 1\) in \(\Pi _ 1(B)\) and a point \(e _ 0 \in E\) such that \(p(e _ 0) = b _ 0\), there is a unique lifting \(\tilde{f} : e _ 0 \rightarrow e _ 1\) in \(\Pi _ 1(E)\) for which \(p _ \ast(\tilde{f}) = f\).
Proof. For covering maps, there is a unique path lifting. Also, two homotopic paths lift to homotopic paths. Hence the lifting is unique. ▨
We use this algebraic property as an analogue of covering maps.
Definition 6. A covering map is a groupoid morphism \(F : \mathcal{E} \rightarrow \mathcal{B}\) which satisfies the unique lifting property. The groupoid \(\mathcal{E}\) is called a covering groupoid of \(\mathcal{B}\) in this case.
Then we can restate the above theorem as the following.
Theorem 7. If \(p : E \rightarrow B\) is a covering map between topological spaces, then the induced map \(p _ \ast : \Pi _ 1(E) \rightarrow \Pi _ 1(B)\) is also a covering map between groupoids.
We see that a simply connected space corresponds to a groupoid in which every \(\hom(x,y)\) has a single element. For some nice topological spaces, we know that there exists a simply connected covering space, namely the universal covering space. We also have an analogue for this for groupoids.
Definition 8. A universal covering groupoid is a covering groupoid for which every \(\hom(x,y)\) has exactly one element.
Theorem 9 (Existence of a universal covering). For any groupoid \(\mathcal{B}\), there exists a universal covering groupoid \(\mathcal{E}\).
Proof. The proof is almost the same as the usual construction of universal covering spaces. We first pick an object \(b _ 0 \in \mathrm{ob}(\mathcal{B})\) and consider all the morphisms (paths) starting from \(b _ 0\). These should be the objects (points) of \(\mathcal{E}\). Project \(\mathrm{ob}(\mathcal{E})\) to \(\mathrm{ob}(\mathcal{B})\) canonically by letting \(\hom(b _ 0, b _ 1) \mapsto b _ 1\). Also for \(f _ 1 : b _ 0 \rightarrow b _ 1\) and \(f _ 2 : b _ 0 \rightarrow b _ 2\), just project \(\hom(f _ 1, f _ 2) \mapsto f _ 2 f _ 1^{-1}\). It is not hard to verify that this map is indeed a covering map. ▨
The great thing about this theorem is that there is no condition imposed on \(\mathcal{B}\) at all! Because we need not care about open sets, the whole picture becomes much simpler when we move from topological spaces to groupoids. There is a groupoid version of the general lifting lemma as well. Using this theorem, we get the universal property of the universal covering as a corollary.
Theorem 10 (General lifting lemma). Let \(p : \mathcal{E} \rightarrow \mathcal{B}\) be a covering morphism, and let \(p(e _ 0) = b _ 0\). Let \(f : \mathcal{Y} \rightarrow \mathcal{B}\) be a morphism with \(f(y _ 0) = b _ 0\) where \(\mathcal{Y}\) is a path connected groupoid. The map \(f\) can be lifted to a map \(\tilde{f} : \mathcal{Y} \rightarrow \mathcal{E}\) such that \(\tilde{f}(y _ 0) = e _ 0\) if and only if \[ \displaystyle f( \hom _ \mathcal{Y}(y _ 0, y _ 0)) \subset p( \hom _ \mathcal{E}(e _ 0, e _ 0)).\] Furthermore, if such a lifting exists, it is unique.
Proof. Clearly, if such a map exists, then \[ \displaystyle f(\hom _ {\mathcal{Y}}(y _ 0, y _ 0)) = p(\tilde{f}(\hom _ {\mathcal{Y}}(y _ 0, y _ 0))) \subset p( \hom _ \mathcal{E}(e _ 0, e _ 0)). \] We now prove the other direction. Suppose that the inclusion is true. Since \(\mathcal{Y}\) is connected, for each \(y \in \mathcal{Y}\) there is a morphism (path) \(\gamma : y _ 0 \rightarrow y\). The projection \(f(\gamma)\) will be some morphism \(b _ 0 \rightarrow b\) where \(b = p(y)\), and by the unique lifting property, will lift uniquely to some \(\tilde{\gamma} : e _ 0 \rightarrow e\) in \(\mathcal{E}\). Since such a path \(\tilde{\gamma}\) for which \(f(\gamma) = p(\tilde{\gamma})\) is unique, we see that any \(\tilde{f}\) for which \(f = p \circ \tilde{f}\) should map \(\gamma\) to \(\tilde{\gamma}\). Hence it should map \(y\) to \(e\). We should check that this map which sends \(y\) to \(e\) is well-defined. Suppose that there are two arrows \(\gamma _ 1, \gamma _ 2 : y _ 0 \rightarrow y\). Then using \(\gamma _ 1\) and \(\gamma _ 2\), we can map \(y\) to two objects in \(\mathcal{E}\). Suppose that we got the arrow \( \tilde{\gamma} _ 1 : e _ 0 \rightarrow e \) after lifting \( f(\gamma _ 1) \). Since \(\gamma _ 1^{-1} \gamma _ 2\) is in \(\hom _ \mathcal{Y}(y _ 0, y _ 0)\) we see that \[ \displaystyle f(\gamma _ 1)^{-1} f(\gamma _ 2) = f(\gamma _ 1^{-1} \gamma _ 2) \in f(\hom _ \mathcal{Y}(y _ 0, y _ 0)) = p(\hom _ \mathcal{E}(e _ 0, e _ 0)).\] Hence there is a lifting of \(\gamma _ 1^{-1} \gamma _ 2\) into \(\mathcal{E}\) as a arrow \(\tilde{\gamma} _ 1^{-1} \tilde{\gamma} _ 2 : e _ 0 \rightarrow e _ 0\). Hence if we compose the two arrows, we see that \(f(\gamma _ 2)\) lifts to an arrow \(\tilde{\gamma} _ 2 : e _ 0 \rightarrow e\). Thus given a \(y\) we see that there is a unique \(e\). One can also check that the arrows of \(\mathcal{Y}\) can be lifted so that \(\tilde{f}\) actually is a functor. ▨
Corollary 11. Let \(\mathcal{E}\) be a universal cover of the groupoid \(\mathcal{B}\) and let \(p : \mathcal{E} \rightarrow \mathcal{B}\) be the corresponding covering map. Let \(r : \mathcal{Y} \rightarrow \mathcal{B}\) any covering map. Then there exist s a unique covering map \(q : \mathcal{E} \rightarrow \mathcal{Y}\) for which \(r \circ q = p\).
Corollary 12. If \(\mathcal{E}\) and \(\mathcal{E}^\prime\) are both universal covering spaces and \(p : \mathcal{E} \rightarrow \mathcal{B}\) and \(p^\prime : \mathcal{E}^\prime \rightarrow \mathcal{B}\) are the associated covering maps, then there is a unique map \(q : \mathcal{E} \rightarrow \mathcal{E}^\prime\) such that \(p = p^\prime \circ q\).
3. Deck transformations#
We now discuss deck transformations.
Definition 13. A deck transformation (also called a deck transformation) of a covering map \(p : E \rightarrow B\) is a homomorphism \(h : E \rightarrow E\) for which the following diagram commutes.
Again, there is an analogue for groupoids.
Definition 14. A deck transformation of a covering map \(p : \mathcal{E} \rightarrow \mathcal{B}\) for groupoids is an isomorphism \(h : \mathcal{E} \rightarrow \mathcal{E}\) for which the following diagram commutes.
Given a covering map \(p\) and a deck transformation \(h\) between topological spaces, we can canonically obtain a deck transformation between their fundamental groupoids.
Theorem 15. Suppose that both \(E\) and \(B\) are path connected and locally path connected. Then for any deck transformation \(g : \Pi _ 1(E) \rightarrow \Pi _ 1(E)\) there is a unique deck transformation \(h : E \rightarrow E\) for which \(g = h _ \ast\). In other words, the inducing map \(\bullet _ \ast : \mathcal{C}(E, p, B) \rightarrow \mathrm{Aut} _ {\Pi _ 1(B)}( \Pi _ 1(E), p _ \ast)\) which maps \(h \mapsto h _ \ast\) is an isomorphism.
Proof. As mentioned earlier, given a deck transformation \(g : \Pi _ 1(E) \rightarrow \Pi _ 1(E)\), there can be at most one \(h : E \rightarrow E\) for which \(g = h _ \ast\). This is because the morphism \(g\) contains all the information about how the objects of \(\Pi _ 1(E)\) maps to the objects of \(\Pi _ 1(E)\). For \(g = h _ \ast\) to be satisfied, the map \(h\) has to be \(g\) restricted on the objects of \(\Pi _ 1(E)\), which is all the points of \(E\). We now prove that such \(h\) is a homeomorphism. From the fact that \(g\) is an isomorphism, it follows that \(h\) is a bijection. If \(h\) be a local homeomorphism, we will be able to conclude that \(h\) is a homeomorphism. It turns out that \(h\) sends one copy of a evenly covered open set to another copy and hence is a local homeomorphism. ▨
This theorem shows that the characterization of deck transformations for topological spaces is (in most cases) equivalent to those for groupoids. Instead of thinking about topological spaces, we move our whole attention to their fundamental groupoids.
4. The fundamental group#
We define the fundamental group in the following way as in BJ01.
Definition 16. Let \(\mathcal{B}\) be a groupoid. We define the Chevalley fundamental group \(\pi _ 1(\mathcal{B})\) as the group \(\mathrm{Aut} _ \mathcal{B}(\mathcal{E}, p)\) where \(\mathcal{E}\) is the universal cover of \(\mathcal{B}\) and \(p : \mathcal{E} \rightarrow \mathcal{B}\) is the corresponding map. If \(\mathcal{B} = \Pi _ 1(B)\) for some topological space \(B\), then we also denote \(\pi _ 1(B) = \pi _ 1(\mathcal{B}) = \mathrm{Aut} _ \mathcal{B}(\mathcal{E}, p)\).
Surprisingly it turns out that the Chevalley fundamental group is isomorphic to the ordinary fundamental group.
Theorem 17. Let \(\mathcal{B}\) be a groupoid, and let \(p : \mathcal{E} \rightarrow \mathcal{B}\) be the universal covering map. Given an object \(e _ 0 \in \mathcal{E}\), there is a canonical isomorphism \(f _ {e _ 0} : \pi _ 1(\mathcal{E}) \rightarrow \hom _ \mathcal{B}(b _ 0, b _ 0)\) where \(b _ 0 = p(e _ 0)\).
Proof. Consider any \(h \in \pi _ 1(\mathcal{E})\). Because \(\mathcal{E}\) is universal, there is a unique element in \(\hom _ \mathcal{E}(h(e _ 0), e _ 0)\). Consider the projection of this unique arrow \(\gamma _ h : h(e _ 0) \rightarrow e _ 0\) to \(\mathcal{B}\). The new arrow \(p(\gamma _ h)\) should be inside \(\hom _ \mathcal{B}(b _ 0, b _ 0)\). Associate \(h\) to \(p(\gamma _ h)\) by \[ \displaystyle f _ {e _ 0} : h \mapsto p(\gamma _ h).\] By Theorem 10 this map \(f _ {e _ 0}\) should be bijective. Also one can check that it is a homomorphisms between groups. ▨
Because the fundamental group depends on the choice of the base point while \(\pi _ 1(B)\) does not, the isomorphism should necessarily depend on a choice of some kind. This is why we need to pick an object to get the isomorphism.
Remark 1. It is worth noting that we benefit from the fact that the universal cover always exists for groupoids. If we had defined the Chevalley fundamental group as the group of deck transformations \(\mathcal{C}(E, p, B)\), where \(E\) is a universal covering space, it would not have been applicable to topological spaces \(B\) that does not admit a universal covering space.
5. Description of deck transformations#
We look at one application of our new definition of the fundamental group. Let us try to figure out what the deck transformation group is in general. Let \(p : \mathcal{E} \rightarrow \mathcal{E}\) a covering morphism, and let \(\mathcal{X}\) be the universal covering groupoid of \(\mathcal{B}\) with the morphism \(r : \mathcal{X} \rightarrow \mathcal{B}\). Then by Theorem 10 there is a covering morphism \(q : \mathcal{X} \rightarrow \mathcal{B}\).
A convenient way is to think such \(h\) as a morphism whose projection to \(\mathcal{E}\) is well defined. Clearly such morphisms \(h\) are closed under composition, because we can paste another diagram on the right side and get the same diagram. Let us denote this group by \(G\). (This notation is bad, but we shall use it throughout this section.) Because \(p \circ q \circ h = p \circ q\), we see that \(G\) is a subgroup of \(\mathrm{Aut} _ \mathcal{B}(\mathcal{X})\). Also this group \(G\) contains \(\mathrm{Aut} _ \mathcal{E}(\mathcal{X})\) because every \(h \in G\) can be projected to the identity map on \(\mathcal{E}\). Hence we get a canonical embedding \[ \mathrm{Aut} _ \mathcal{E}(\mathcal{X}) \hookrightarrow G. \] Given an \(h\), we get a map \(g : \mathcal{E} \rightarrow \mathcal{E}\). This projected morphism \(g\) is unique, because \(q\) is surjective. Also, given any morphism \(g : \mathcal{E} \rightarrow \mathcal{E}\), there is a lifting \(h : \mathcal{X} \rightarrow \mathcal{X}\), because by Theorem 10 we can always lift \(g \circ q : \mathcal{X} \rightarrow \mathcal{E}\) to \(h\). Therefore we get a canonical surjection \[ \displaystyle G \twoheadrightarrow \mathrm{Aut} _ \mathcal{B}(\mathcal{E}). \]
Theorem 18. The sequence of group homomorphisms \[ \displaystyle (1) \to \mathrm{Aut} _ \mathcal{E}(\mathcal{X}) \hookrightarrow G \twoheadrightarrow \mathrm{Aut} _ \mathcal{B}(\mathcal{E}) \to (1) \] is a short exact sequence.
Proof. Because we know that the second map is injective and the third map is surjective, we need only check that the kernel and the image agree in \(G\). The image of the second map is the subgroup of maps whose projection is the identity. Also, the kernel of the third map is the subgroup of \(h\)s for which the \(g\) is the identity. These two statements are equivalent, and hence the sequence is exact. ▨
It follows that \[ \displaystyle \mathrm{Aut} _ \mathcal{B}(\mathcal{E}) \simeq G / \mathrm{Aut} _ \mathcal{E}(\mathcal{X}).\] Let us now find out what \(G\) exactly is. After some playing around, one gets the following lemma.
Lemma 19. Denote \(G _ 0 = \mathrm{Aut} _ \mathcal{B}(\mathcal{X})\) and \(H _ 0 = \mathrm{Aut} _ \mathcal{E}(\mathcal{X})\). We have the inclusion \(H _ 0 \subset G \subset G _ 0\). The group \(G\) is actually the normalizer of \(H _ 0\) in \(G _ 0\), i.e., \[ \displaystyle G = N(H _ 0) = { h \in G _ 0 : h H _ 0 h^{-1} = H _ 0}. \]
Corollary 20. There is a canonical isomorphism \[ \displaystyle \mathrm{Aut} _ \mathcal{B}(\mathcal{E}) \simeq N(\pi _ 1(\mathcal{E})) / \pi _ 1(\mathcal{E}) \] inside \(\pi _ 1(\mathcal{B})\).
If we assume that both \(B\) and \(E\) are path connected and locally path connected, then from Theorem 15 we immediately get \[ \displaystyle \mathcal{C}(E, p, B) \simeq N(\pi _ 1(E)) / \pi _ 1(E) \] where \(N(\pi _ 1(E))\) is the normalizer of \(\pi _ 1(E)\) inside \(\pi _ 1(B)\).
References#
[BJ01] Francis Borceux and George Janelidze, Galois theories, Cambridge Studies in Advanced Mathematics, vol. 72, Cambridge University Press, Cambridge, 2001.
[Mun00] James Munkres, Topology, 2nd ed., Prentice Hall, Upper Saddle River, NJ, 2000.