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An orientation of a topological manifold chooses a generator of every local homology group $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$, compatibly on coordinate neighbourhoods. Each choice is a local orientation of a manifold. Compatibility means the generators are locally restrictions of the same relative orientation class. It can be checked by choosing oriented coordinate balls and requiring agreement wherever their local classes overlap.

For a connected closed manifold, an orientation gives a unique fundamental class in $H_n(M;\mathbb{Z})$ whose restriction at each point is the chosen local generator. In an oriented triangulation this class is the sum of compatibly signed top simplices: interior faces cancel in opposite pairs, while the local coefficient in every top simplex is one. Reversing every local generator negates the fundamental class.

With the usual second-countability convention, an $n$-manifold is a Hausdorff second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$. A local orientation of a manifold at $x$ is a choice of generator ofExcision and a coordinate ball identify this relative group with $H_n({\mathbb{R}}^n,{\mathbb{R}}^n\setminus \{0\})$. An orientation of a topological manifold is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with orientation-preserving transition maps.

For a connected closed oriented manifold, its fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to every chosen local generator. Here closed means compact without boundary. Poincare duality states that cap product with this class is an isomorphismOver rational coefficients it gives $b_k=b_{n-k}$. ThereforeWhen $n$ is odd this says $\chi =-\chi$, so $\chi (M)=0$. Torsion does not contribute to the Betti numbers or this Euler characteristic.

Remove the ball interiors. Each remaining piece inherits an orientation of a manifold and hence a boundary orientation. Choose the orientation on the second piece, reversing its original choice if necessary, so that the gluing map identifies these boundary orientations oppositely. Its two possible orientation signs are accommodated by this choice.

Near a glued point, the two half-space collar charts combine to give a full Euclidean chart. The opposite induced boundary orientations make the two interior orientations agree in this chart. Away from the seam, keep the orientations already chosen on the two pieces. These fit together globally, so the connected sum is orientable. This does not require that an arbitrarily chosen gluing preserve both initially specified orientations unchanged.

A closed manifold is a compact manifold without boundary. Here $M$ is Hausdorff and every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$; the usual countability condition is included in the manifold convention. Compactness also gives a finite atlas and hence second countability.

A local orientation at $x$ is a choice of generator of $H_n(M,M\setminus \{x\};\mathbb{Z})\cong \mathbb{Z}$. An orientation of a manifold is a compatible choice of these generators as $x$ varies: locally they are induced by the same coordinate-ball orientation, or equivalently by a local relative homology class. For an oriented closed manifold, the fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to the chosen generator at every point. On disconnected closed manifolds this is the sum of the fundamental classes of their finitely many components.

For a triangulated or Delta complex description, orient the top-dimensional simplices compatibly with the local orientation and sum them with their corresponding signs. Each codimension-one face appears twice with opposite induced orientations, so the sum is a cycle. At a point inside any top simplex its image is the prescribed local generator, so this cycle represents $[M]$. In the connected case $H_n(M)=\mathbb{Z}$ generated by this class.

The homology conclusions below use the opening closed-manifold hypothesis for $M_1,M_2$. That hypothesis is necessary if their later description is read independently: removing a ball from ${\mathbb{R}}^n$ for $n\ge 2$ creates a nonzero $(n-1)$st homology group, so the claimed isomorphism would fail without closedness.