Existence of the fundamental class

The blog might be back, let's see how consistent I stay with it. The goal of this post is to present the details of what I think is a nice proof (from Hatcher) which characterizes the top-dimensional homology of closed connected (topological) \(n\) -manifolds.

First, briefly recall that we use the notation \(H_n(M \ | \ x; R)\) to denote the relative homology \(H_n(M, M - \{x\}; R) \simeq R\) (by the locally Euclidean property of \(M\), excision, the long exact sequence of homology). We also use \(H_n(M | U; R) = H_n(M, M - U; R)\) when, more generally, we are removing an open set \(U\). \(M_R\) denotes the space of \(\alpha_x \in H_n(M \ | \ x; R)\), for each \(x \in M\): this space is naturally a smooth manifold over \(M\) where the projection \(\pi : M_R \to M\) sending \(\alpha_x\) to its corresponding \(x \in M\) is a covering map. When we omit \(R\) and simply use \(H_n(M | x)\), we are implicitly setting \(R = \mathbb{Z}\). Note that we have a canonical isomorphism \(H_n(M | x) \otimes R \simeq H_n(M | x; R)\) obtained from the obvious isomorphism of \(R\) -modules at the level of relative chains \(C_n(M, M - \{x\}) \otimes R \simeq C_n(M, M - \{x\}; R)\). It follows that if we fix \(r \in R\), then we obtain a subcovering space \(M_r \subset M_R\) consisting of the points \(\pm \mu_x \otimes r\). If \(r = -r\) in \(R\), then these points are equal, and the subcovering space is just a copy of \(M\). Otherwise, we get a copy of \(\widetilde{M}\): the usual two-sheeted orientation cover for \(M\). It then follows that \(M_R\) is the union of all the \(M_r\), where clearly \(M_a = M_{b}\) if and only if \(a = \pm b\).

An \(R\) -orientation for \(M\) is a choice of generator/unit \(\mu_x \in H_n(M | x; R)\) (local orientation) for each \(x \in M\) such that for any such \(x\), there exists a coordinate ball \(B\) around \(x\) such that \(\mu_y\) for all \(y \in B\) is the image of some \(\mu_B \in H_n(M | B; R) \simeq R\) under the natural inclusion \(H_n(M | B; R) \to H_n(M | x; R)\). The manifold \(M\) is said to be \(R\) -orientable if it can be endowed with an orientation. Recall the following result:

Claim. A connected manifold \(M\) is \(\mathbb{Z}\) -orientable if and only if \(\widetilde{M}\) has two connected components.

Proof. If \(\widetilde{M}\) has two connected components, each is homeomorphic to \(M\), and since \(\widetilde{M}\) is orientable, \(M\) must be as well. Conversely, if \(M\) is orientable, it is easy to check that it has precisely two orientations. We claim that each of the images of these individual orientations in \(\widetilde{M}\) determine two connected components. Indeed, these images (call them \(M_{\pm}\)) must cover all of \(\widetilde{M}\), are disjoint, and are connected as the continuous image of \(M\) which is connected. In fact, since a cover is a local homeomorphism, and the restricted projections \(M_{\pm} \to M\) are bijective, each \(M_{\pm}\) is necessarily homeomorphic to \(M\), so each is clopen. It follows that \(M_{+}\) and \(M_{-}\) must be individual connected components. \(\blacksquare\)

From the above characterization of \(M_R\) it is easy to prove the following results:

Claim. If connected \(M\) is \(\mathbb{Z}\) -orientable, it is \(R\) -orientable for any \(R\). If \(M\) is not \(\mathbb{Z}\) -orientable, it is \(R\) -orientable if and only if \(2 = 0\) in \(R\).

Proof. If \(M\) is \(\mathbb{Z}\) -orientable, then \(M_R\) always consists of a disjoint collection of sheets each homeomorphic to \(M\). In particular, such a homeomorphism \(M_1 \to M\) gives rise to an \(R\) -orientation. If \(M\) is not \(\mathbb{Z}\) -orientable and \(2 \neq 0\), then \(-r \neq r\) for any unit \(r \in R\), hence if there did exist a section \(M \to M_r\) for some unit \(r\), then it would give rise to a section \(M \to \widetilde{M}\), which is a contradiction. Conversely, if \(2 = 0\), then \(M_r \simeq M\) for any unit \(r\), and we can easily write down a section, which is precisely an \(R\) -orientation. \(\blacksquare\)

From here, we can move to the main result of this post:

Theorem 1. Let \(M\) be closed and connected. If \(M\) is \(R\) -orientable, then the natural map \(H_n(M; R) \to H_n(M | x; R) \simeq R\) is an isomorphism for all \(x \in M\). If \(M\) is not \(R\) -orientable, then the same map is injective, with image consisting of all elements \(\mu_x \in H_n(M | x; R)\) such that \(2\mu_x = 0\). In addition, \(H_i(M; R) = 0\) for all \(i > n = \dim(M)\).

To prove this result, we will prove the following lemma:

Lemma 1. If \(M\) is an \(n\) -manifold and \(A \subset M\) is a compact subset, then given a section \(x \mapsto \alpha(x)\) of the cover \(M_R \to M\), there is a unique class \(\Phi(\alpha) \in H_n(M | A; R)\) whose image is \(\alpha(x) \in H_n(M | x; R)\) for all \(x \in A\), under the usual induced map. Moreover, \(H_i(M | A; R) = 0\) for \(i > n\).

Let us first see how Lemma 1 will imply Theorem 1:

Proof of Theorem 1. When \(M\) is closed, it is compact, so we can set \(A = M\) to immediately get \(H_i(M; R) = H_i(M | M; R) = 0\) for \(i > n\). In addition, if we let \(\Gamma_R(M)\) denote the group of smooth sections of \(M \to M_R\), then we have map \(\Phi : \Gamma_R(M) \to H_n(M; R)\) with \(\alpha \mapsto \Phi(\alpha)\) and \(\Psi : H_n(M; R) \to \Gamma_R(M)\) with \(\Psi(\beta)(x)\) defined to be the image of \(\beta\) in \(H_n(M | x; R)\) under the map induced by inclusion. Clearly, \(\Psi(\beta)(x)\) projects to \(x\) under \(M_R \to M\). Moreover, the map \(\Psi(\beta)\) is continuous. Let \(U(\mu_B)\) be an open set in \(M_R\) consisting of points of \(H_n(M | x; R)\) for \(x \in B\) which are images of \(\mu_B \in H_n(M | B; R)\). Note that \(\Psi(\beta)(x) \in U(\mu_B)\) if and only if \(x \in B\) and \(\beta_x = (\beta_B)_x = (\mu_B)_x\). Of course, the map \(H_n(M | B; R) \to H_n(M | x; R)\) is an isomorphism, so we must have \(\beta_B = \mu_B\). Thus, \(\Psi(\beta)^{-1}(U(\mu_B))\) is either empty, or \(B\). Thus, \(\Psi(\beta)\) is, in fact, a section.

Lemma 1 will imply that \(\Psi(\Phi(\alpha))(x) = \alpha(x)\) for all \(x \in A\), so \(\Psi \circ \Phi = \text{id}\). In addition, it is clear that any \(\beta \in H_n(M; R)\) is a class such that the image of \(\beta\) in \(H_n(M | x; R)\) is \(\Psi(\beta)(x)\) for each \(x \in A\), so \(\Phi(\Psi(\beta)) = \beta\) by uniqueness, and \(\Phi \circ \Psi = \text{id}\). Thus, we have bijections between the two groups. To see that these maps are homomorphisms, simply note that the image of a sum of classes \(\beta_1 + \beta_2\) in \(H_n(M | x; R)\) is equal to the sum of the images, for all \(x \in A\), so \(\Psi(\beta_1 + \beta_2)(x) = \Psi(\beta_1)(x) + \Psi(\beta_2)(x)\), and we thus have \(\Psi(\beta_1 + \beta_2) = \Psi(\beta_1) + \Psi(\beta_2)\). Similarly, given sections \(\alpha_1, \alpha_2\),

\begin{equation} \Phi(\alpha_1 + \alpha_2) = \Phi(\Psi(\Phi(\alpha_1)) + \Psi(\Phi(\alpha_2))) = (\Phi \circ \Psi)(\Phi(\alpha_1) + \Phi(\alpha_2)) = \Phi(\alpha_1) + \Phi(\alpha_2) \end{equation}

so we in fact have an isomorphism of these groups, \(H_n(M; R) \simeq \Gamma_R(M)\). From here, note that if \(M\) is connected, then any section of a covering map \(M_R \to M\) is determined uniquely by its value at a single point. In particular, it is easy to see from the structure of \(M_R \to M\) discussed above as a disjoint union of sheets of the form \(M_r\) that when \(M\) is \(R\) -orientable, every element of \(H_n(M | x; R)\) for some \(x \in M\) will determine a section, whereas when \(M\) is not \(R\) -orientable, only the \(r\) corresponding to sheets where \(r = -r\) will determine sections. This gives us the remaining implications of Theorem 1. \(\blacksquare\)

Let us now finish the blog post by proving Lemma 1:

Proof of Lemma 1. The idea with this proof is to reduce the problem a few times to easier and easier cases. Going forward, I will suppress the notation of putting \(R\) in \(H_n(M | A; R)\), and just denote this group by \(H_n(M | A)\), to save space.

1. First, note that if the claim holds for compact sets \(A\), \(B\), \(A \cap B\), then it holds for \(A \cup B\). A section \(\alpha \in \Gamma_R(M)\) will restrict to sections on \(A, B, A \cap B\). Each of these sections will then give rise to unique classes \(\alpha_A \in H_n(M | A)\), \(\alpha_B \in H_n(M | B)\) \(\alpha_{A \cap B} \in H_n(M | A \cap B)\), which have image equal to \(\alpha(x)\) in \(H_n(M|x)\) for \(x\) in \(A\), \(B\), and \(A \cap B\) respectively. Clearly, the images of \(\alpha_A\) and \(\alpha_B\) in \(H_n(M | A \cap B)\) satisfy the property which defines \(\alpha_{A \cap B}\), namely that its image is \(\alpha(x)\) in \(H_n(M | x)\) for all \(x \in A \cap B\), so by uniqueness of the class, the images of \(\alpha_A\) and \(\alpha_B\) in \(H_n(M | A \cap B)\) are equal to \(\alpha_{A \cap B}\). We have a relative Mayer-Vietoris sequence of the form:

   \begin{equation} 0 \longrightarrow H_n(M | A \cup B) \overset{\Phi}{\longrightarrow} H_n(M | A) \oplus H_n(M | B) \overset{\Psi}{\longrightarrow} H_n(M | A \cap B) \end{equation}

   where the zero of the left comes from the assumption that \(H_{i}(M | A \cap B; R) = 0\) for any \(i > n\). Automatically, if we extend the Mayer-Vietoris sequence farther to the left, we will also have \(H_i(M | A \cup B) = 0\) for all \(i > n\) from the assumptions. The Mayer-Vietoris maps are the usual ones, so exactness implies that \(\Phi(\alpha_{A \cup B}) = (\alpha_A, -\alpha_B)\) for some \(\alpha_{A \cup B} \in H_n(M | A \cup B)\). The map \(\Phi\), in particular, is taking the image of \(\alpha_{A \cup B}\) in \(H_n(M|A)\) and \(H_n(M|B)\) (and then flipping the sign of the \(H_n(M|B)\) entry), so it follows that taking the image of \(\alpha_{A \cup B}\) in \(H_n(M | x)\) is \(\alpha(x)\), as \(\alpha_A\) and \(\alpha_B\) have this property. To see that \(\alpha_{A \cup B}\) is unique, note that if \(\alpha_{A \cup B}\) has image \(0\) in \(H_n(M | x)\) for all \(x\), then its images in \(H_n(M | A)\) and \(H_n(M | B)\) respectively must also both have images (which we call \(\alpha'_A\) and \(\alpha'_B\)) equal to \(0\) when we further take the image in \(H_n(M | x)\). Thus, by uniqueness of \(\alpha'_A\) and \(\alpha'_B\), from the assumption on \(A\) and \(B\), we must have \(\alpha'_A = \alpha_B' = 0\), so by injectivity of \(\Phi\) in the Mayer-Vietoris sequence, \(\alpha_{A \cup B} = 0\) as well. So, the map assigning a section to \(\alpha_{A \cup B}\) is injective, and we have our desired uniqueness.

2. From here, we note a that a compact set \(A\) can be written as a finite union of compact sets \(A_1 \cup \cdots \cup A_m\) each contained in a homeomorphic copy of \(\mathbb{R}^n \subset M\) (we just take a finite cover of \(A\) by Euclidean balls whose closure in contained in a slightly larger Euclidean ball). We can apply what we showed above to \(A_1 \cup \cdots \cup A_{m - 1}\) and \(A_{m - 1}\): namely, the intersection of these sets is \((A_1 \cap A_m) \cup \cdots \cup (A_{m-1} \cap A_m)\) which is a union of \(m - 1\) compact sets each contained an a homeomorphic copy of \(\mathbb{R}^n \subset M\), hence using the previous paragraph, by induction on \(M\), we have reduced proving the claim to proving the case of \(m = 1\). So, we can assume WLOG that our compact set \(A\) is contained in a homeomorphic copy of \(\mathbb{R}^n\) in \(M\). In fact, without loss of generality, we can just work in \(\mathbb{R}^n\), assuming \(A\) is a compact subset of \(\mathbb{R}^n\), as the claim we are trying to prove only involves \(H_n(M | A)\), which by excision is the same as \(H_n(U | A)\) for some Euclidean ball \(U\) around \(A\) homeomorphic to \(\mathbb{R}^n\).

3. When \(A \subset \mathbb{R}^n\) is a union of convex compact sets \(A_1, \dots, A_m\) in \(\mathbb{R}^n\), we know that the intersection of convex compact sets will be convex compact, hence by the same inductive logic as the previous paragraph, we can reduce to the case that \(A\) is convex compact in \(\mathbb{R}^n\). But in this case, the result becomes obvious: the inclusion maps \(H_i(\mathbb{R}^n | A) \to H_i(\mathbb{R}^n | x)\) are isomorphisms because \(\mathbb{R}^n - \{x\}\) deformation retracts to \(\mathbb{R}^n - A\) via straight-line homotopy (as \(A\) is convex). Thus, inclusion \(\iota : \mathbb{R}^n - A \to \mathbb{R}^n - \{x\}\) induces an isomorphism in homology, and we can apply naturality and the five lemma to the long exact sequences in homology. Therefore, if \(\alpha \in \Gamma_R(M)\) is our section, then we can just set \(\Phi(\alpha)\) to be the inverse image of \(\alpha(x)\) under this isomorphism, for any \(x \in A\). A section \(\alpha : A \to M_R\) is a lift of the inclusion \(A \to M\) to the cover. Since the section \(y \mapsto (\Phi(\alpha))_y \in H_n(M | y)\) and \(y \mapsto \alpha(y)\) agree at point \(y = x\), it follows that they must be equal on all of \(A\) (as \(A\) is convex, hence connected), so \(\Phi(\alpha)\) has the desired property.

4. The final case we must consider is when \(A \subset \mathbb{R}^n\) is an arbitrary compact subset. Let \(\beta \in H_i(\mathbb{R}^n | A)\) be represented by a relative cycle \(z\) in \(\mathbb{R}^n\), with \(\partial z \in \mathbb{R}^n - A\), let \(C\) be the union of the images of the singular simplices in \(\partial z\). Since \(C\) is compact (finite union of images of compact sets), it has positive distance \(\delta\) from \(A\) in \(\mathbb{R}^n\) (this is a basic topological fact). We cover \(A\) by finitely many closed balls of radius less than \(\delta\) centred at points of \(A\), let \(K\) be the union of these balls (a finite union of convex compact sets), it is disjoint from \(C\). It follows that the relative cycle \(z\) in fact has its boundary \(\partial z\) in \(\mathbb{R}^n - K\). Therefore, this relative cycle defines an element \(\beta_K \in H_i(\mathbb{R}^n | K)\), which by definition, maps to \(\beta \in H_i(\mathbb{R}^n | A)\) under the inclusion map. Set \(i = n\) going forward: we will show that if \(\beta_x \in H_n(\mathbb{R}^n | x)\) is \(0\) for all \(x \in A\), then \(\beta = 0\). As we noted earlier, \(K\) is a union of balls \(B\) meeting \(A\), and each map \(H_n(\mathbb{R}^n | B) \to H_n(\mathbb{R}^n | x)\) is an isomorphism for all \(x \in B\). It follows that \(\beta_K\) maps to \(0\) in \(H_n(M | x)\) for each \(x \in K\): the reason being that \((\beta_K) \mapsto (\beta_K)_x\) factors through \((\beta_K) \mapsto (\beta_K)_B \mapsto (\beta_K)_x\), where \(B\) some ball containing \(x\) and \((\beta_K)_B\) is the image in \(H_n(\mathbb{R}^n | B)\). But each such ball contains \(y \in A\) as well, and taking the image of \(\beta_K\) in \(H_n(\mathbb{R}^n | y)\) is the same as taking the image of \(\beta\) in \(H_n(\mathbb{R}^n | y)\), which we assumed will yield \(0\). Hence, since \(H_n(\mathbb{R}^n | B) \to H_n(\mathbb{R}^n | y)\) is an isomorphism, we must have \((\beta_K)_B = 0\), hence \((\beta_K)_x = 0\) as well. Since \(K\) is a finite union of convex compact sets, we can apply the result of the previous paragraph (particularly, uniqueness) to conclude that \(\beta_K = 0\), hence its image \(\beta\) in \(H_n(\mathbb{R}^n | A)\) is also \(0\).

   Thus, to conclude, given a section \(\alpha \in \Gamma_R(M)\) with \(A \subset \mathbb{R}^n\), suppose \(\Phi(\alpha) \in H_n(\mathbb{R}^n | A)\) such that \(\Phi(\alpha)_x = \alpha(x)\) for all \(x \in A\) does in fact exist. If \(\beta\) were another class such that \(\beta_x = \alpha(x)\) for all \(x\), then the image of \(\Phi(\alpha) - \beta\) in \(H_n(\mathbb{R}^n | x)\) is \(0\) for all \(x\), so from (4), \(\Phi(\alpha) = \beta\), and we have the uniqueness claim. To prove existence, we can just take the image of the \(\Phi(\alpha) \in H_n(\mathbb{R}^n | K)\) produced for the union of convex compact sets \(K\) constructed in (3), in \(H_n(\mathbb{R}^n | A)\), and let this be the \(\Phi(\alpha)\) for \(A\). This completes the proof.