Let $M$ a compact, connected $m$-manifold. Assume that there exists a global orientation class, i.e. an $\sigma_M \in H_m(M;\mathbb{Z})$ such that for all $x\in M$ the element $(\rho_{x,M})_{\ast}(\sigma_M)$ is a generator of the local homology group $H_m(M,M\setminus x;\mathbb{Z})\cong \mathbb{Z}.$ Here $(\rho_{x,M})_{\ast}\colon H_m(M;\mathbb{Z})\rightarrow H_m(M,M\setminus x;\mathbb{Z})$ is the morphism induced by the map of pairs $\rho_{x,M}\colon (M,\emptyset)\rightarrow (M,M\setminus x)$. I want to prove that then $H_m(M;\mathbb{Z})\cong \mathbb{Z}.$
First, since no generator of $\mathbb{Z}$ has finite order and since applying a group homomorphism can not increase the order of an element, $\sigma_M$ has infinite order.
Now I want to prove that $\sigma_M$ is a generator of $H_m(M;\mathbb{Z})$. To see this, let $a\in H_m(M;\mathbb{Z})$. Then for all $x\in M$ there exists $k_{x,a}\in \mathbb{Z}$ with $(\rho_{x,M})_{\ast}(a)=k_{x,a}\cdot (\rho_{x,M})_{\ast}(\sigma_M)$, since $(\rho_{x,M})_{\ast}(\sigma_M)$ is a generator.
My notes now read as follows:
As the $(\rho_{x,M})_{\ast}(\sigma_M)$ are all coherent, these $k_{x,a}$ have to be constant, i.e. equal to some $k$, and if we set $b:=a-k\sigma_M$, we see $(\rho_{x,M})_{\ast}(b)=0$ for all $x\in M$. Therefore $b=0$.
How does the sentence marked in bold follow from the fact that the local orientations $(\rho_{x,M})_{\ast}(\sigma_M)$ are all coherent? Put differently, why is $\sigma_M$ a generator?
Edit on the definitions used:
Let $M$ be an $m$-manifold.
- For $x\in M$, a choice of generator $\sigma_x\in H_m(M,M\setminus x)$ is called a local orientation at x.
- A family of local orientations $(\sigma_x)_{x\in M}$ is called coherent if for any $x\in M$ there exists an open neighbourhood $U_x$ of $x$ and an element $\sigma_{U_x}\in H_m(M,M\setminus U_x)$ such that $(\rho_{y,U_x})_{\ast}(\sigma_{U_x})=\sigma_y$ for all $y\in U_x$.
- An orientation on $M$ is a choice of a coherent family of local orientations.
The proof strategy can be summarized as follows: First, we prove the claim for the charts of $M$. Then we transport along paths.
Let $x,y\in M$. Since $M$ is path-connected (any connected manifold is path-connected), there is a path $\gamma\colon [0,1]\rightarrow M$ from $x$ to $y$. For any $z\in \operatorname{im}(\gamma)$, find an open neighbourhood $z\in U_z\cong \mathring{\mathbb{D}}^m$. These $U_z$ form an open cover of the compact set $\operatorname{im}(\gamma)$. We choose a minimal finite subcover $\{U_1,\ldots,U_n\}$ consisting of $n$ elements, say. Next we prove $k_x=k_y$ by induction on the minimal size $n$ of a finite subcover.
Before doing so, I'll simplify notation for better legibility. Namely, set $\sigma_x:= (\rho_{x,M})_{\ast}(\sigma_M)$ and $\sigma_y:= (\rho_{y,M})_{\ast}(\sigma_M)$.
For the base case $n=1$, note that by functoriality of homology $$k_{x}\sigma_x=(\rho_{x,M})_{\ast}(a)= (\rho_{x,U_1})_{\ast}\circ (\rho_{U_1,M})_{\ast}(a)$$ $$k_{y}\sigma_y=(\rho_{y,U_1})_{\ast}\circ (\rho_{U_1,M})_{\ast}(a).$$ Additionally, observe that by excision both $(\rho_{x,U_1})_{\ast}$ and $(\rho_{y,U_1})_{\ast}$ are isomorphisms. Thus, we find $$k_x\cdot (\rho_{x,U_1})_{\ast}^{-1}(\sigma_x)=(\rho_{x,U_1})_{\ast}^{-1}(k_{x}\sigma_x)= (\rho_{y,U_1})_{\ast}^{-1}(k_{y}\sigma_y)= k_y\cdot (\rho_{y,U_1})_{\ast}^{-1}(\sigma_y).$$ Again by functoriality of homology, we have $(\rho_{x,U_1})_{\ast}^{-1}(\sigma_x)= (\rho_{U_1,M})_{\ast}(\sigma_M)= (\rho_{y,U_1})_{\ast}^{-1}(\sigma_y).$ Moreover, since $\sigma_x$ is a generator of $H_m(M,M\setminus x;\mathbb{Z})\cong \mathbb{Z}$ and $(\rho_{x,U_1})_{\ast}$ is an isomorphism, we know that also $(\rho_{x,U_1})_{\ast}^{-1}(\sigma_x)=(\rho_{y,U_1})_{\ast}^{-1}(\sigma_y)$ is a generator of $H_m(M,M\setminus U_1;\mathbb{Z})\cong \mathbb{Z}$. Thus $k_x=k_y$.
Next, assume that the claim is proven for all pairs of points which posses a path between them whose image can be covered by $n-1\geq 1$ open neighbourhoods of the above form. I'll give a kind of parallel transport argument. Let $x,y\in M$ be points with the property that no path between them can be covered by $n-1$ open neighbourhoods of the above form, but for which a path exists which can be covered by $n$ open neighbourhoods of the above form. Pick such a path $\gamma$ and such a minimal open cover $\{U_1,\ldots,U_n\}$ of size $n$. Since the image of $\gamma$ is connected and since $n$ is minimal, there exists $j\neq 1$ such that $U_1\cap U_j$ is non-empty. Pick such a $j$. Similarly, if $n\neq 2$, there exists $i\neq 1$ such that $U_i\cap(U_1\cup U_j)$ is non-empty. We can repeat this procedure and find an order $U_1,U_j,U_i,\ldots$ on the $U_l$'s in the open cover such that each $U_l$ intersects the union of all its predecessors. Without loss of generality assume that the order found is given by the order of the indices, i.e. $j=2$, $i=3$, and so on. Next, note that $\bigcup_{l\leq n-1}U_l$ is path-connected: $U_1\cup U_2$ is a union of two path-connected sets with non-empty intersection, thus it is path connected. Analogously, $(U_1\cup U_2)\cup U_3$ is the union of two path-connected sets with non-empty intersection. And so on (secretly this is an induction proof) … Thus, we can apply the induction hypothesis, and find that $k_{z_1}=k_{z_2}$ for any $z_1,z_2\in\bigcup_{l\leq n-1}U_l$. But $U_n\cap \bigcup_{i\leq n-1}U_i$ is non-empty and $k_x=k_y$ follows.