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Let $f:M\to N$ be a continuous map between oriented $d$-dimensional manifolds, with $d\ge 1$. Suppose $x$ is an isolated point of the fiber over $y=f(x)$. Choose an open neighborhood $U$ with $U\cap f^{-1}(y)=\{x\}$. The map of pairsinduces a map between rank-one degree-$d$ local homology groups. Relative to the local orientation of a manifold, this map is multiplication by an integer, the local degree. Excision makes it independent of the neighborhood. A smooth map between manifolds with invertible derivative at $x$ has local degree $\mathrm{sign}\det D{f}_x$, by the inverse function theorem and the chosen orientations.

Choose a generator $[S^n]$ of ${\tilde{H}}_n(S^n;\mathbb{Z})\cong \mathbb{Z}$. The mapping degree of a continuous map is the integer determined byFor $n\ge 1$ this is ordinary top-dimensional homology; using reduced homology also gives a convention for $n=0$. Domain and target use the same orientation, so reversing the chosen generator in both leaves the integer unchanged. Homotopy invariance of homology makes the mapping degree invariant under homotopy.

We interpret the manifold criterion for a suspension in the category of nonempty topological manifolds without boundary, and “sphere” as homeomorphic to the standard sphere. Let $X$ have dimension $d$. If $X\cong S^d$, an explicit homeomorphism from its suspension to $S^{d+1}$ isIt is a continuous map that is a bijection: the two collapsed ends become the poles, and every other point determines $t$ and $x$ uniquely. The compact-to-Hausdorff continuous bijection theorem makes it a homeomorphism.

Conversely suppose $SX$ is a topological manifold. Its open part $X\times (0,1)$ shows its dimension is $d+1$. At a suspension vertex $v$ there is an open cone neighbourhood $U$ on $X$. It is contractible, while $U\setminus \{v\}$ admits a deformation retraction onto $X$. The long exact sequence in relative homology and excision give the local homology from a link calculationA point of a $(d+1)$-dimensional topological manifold without boundary has local homology $\mathbb{Z}$ in degree $d+1$ and zero in all other degrees. ConsequentlyThus $X$ is an integral homology sphere. This conclusion alone does not identify its homeomorphism type.

If $d\ge 2$, it is also simply connected. Choose a cone neighbourhood $U$ of the vertex and an open coordinate ball $B$ around the vertex inside $U$. By compactness of $X$, sufficiently short entire cone neighbourhoods fit inside any neighbourhood of $v$: the inverse image of that neighbourhood contains $X\times \{0\}$, and a finite subcover supplies one uniform collar length. Choose such a shorter cone $V\subset B$. The inclusions of punctured neighbourhoods factor asThe composite induces an isomorphism of fundamental groups, since both cone punctures are products of $X$ with intervals and their inclusion is a homotopy equivalence. The middle punctured ball has the homotopy type of $S^d$ and is simply connected for $d\ge 2$. Hence the composite is also zero on the fundamental group, forcing ${\pi }_1(X)=0$. This uses the nested neighbourhoods, not an unsupported claim that an arbitrary punctured neighbourhood is a sphere.

The Hurewicz theorem, applied successively to the vanishing lower homology groups, now gives ${\pi }_j(X)=0$ for $j<d$ and ${\pi }_d(X)\cong H_d(X)\cong \mathbb{Z}$. A representative of its generator is a continuous map $S^d\to X$ inducing an isomorphism in every integral homology degree. Topological manifolds have the homotopy type of CW complexes, so the homological Whitehead theorem makes it a homotopy equivalence. We have proved that $X$ is a homotopy sphere.

At this point the topological classification theorem is essential: the topological generalized Poincare theorem says that a closed homotopy sphere is homeomorphic to the standard sphere. Applying it completes the converse for $d\ge 2$. For $d=1$, the local calculation makes $X$ a connected compact one-dimensional manifold without boundary, hence a circle. For $d=0$, a compact zero-dimensional manifold is a finite set, and ${\tilde{H}}_0(X)\cong \mathbb{Z}$ forces exactly two points. Thus it is $S^0$. Altogether,The last classification step is not a consequence of homology alone, and we make no assertion of diffeomorphism with a prescribed smooth structure. If the word manifold allows boundary on both sides, the printed assertion needs this correction: the suspension of the interval is a closed disk. For instance $[x,t]\mapsto ((2x-1)(1-|2t-1|),2t-1)$ identifies it with the filled diamond $|u|+|v|\le 1$, although the interval is not a sphere.

To construct sphere maps of arbitrary integer degree, first use $h_m:S^1\to S^1$, $h_m(z)=z^m$ for any integer $m$, including the constant map when $m=0$. Lifting its argument to the real line gives $\theta \mapsto m\theta$, so the fundamental one-cycle is sent to $m$ times itself. Therefore $\mathrm{deg}{h}_m=m$.

Two cone neighbourhoods cover a suspension; their intersection admits a deformation retraction onto the original space. The reduced Mayer–Vietoris theorem gives a natural suspension isomorphismNaturality means that the suspension of a continuous map induces the same integer multiplier. This proves degree under suspension. Using the standard identification $S(S^k)\cong S^{k+1}$, setThus every integer occurs in every positive sphere dimension, with negative degrees handled just as well as positive ones.

If “every $n$” includes zero, that final printed assertion is false. The generator of ${\tilde{H}}_0(S^0)$ is the difference of its two points. A map of a two-point set is the identity, the transposition, or one of two constant maps, giving preciselyThese degrees of maps of the zero-sphere are the explicit exception to the arbitrary-degree construction.

Choose the intersection point $p=[({\theta }_0,{\theta }_1)]$ as the zero-cell. Let $a$ be the horizontal one-cell $S^1\times \{{\theta }_1\}$ and $b$ the vertical one-cell $\{{\theta }_0\}\times S^1$. The usual CW complex structure on the torus has one two-cell $c$ attached by the word $ab{a}^{-1}{b}^{-1}$, and the extra disk gives a two-cell $d$ attached along $b$ with degree one. Collapsing $a$ leaves one zero-cell $p$, one one-cell $b$, and the two two-cells $c,d$. The cellular boundary formula givesTherefore

The Excision theorem says that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq \mathrm{int}A$, then inclusion inducesIt lets us compute local homology in arbitrarily small neighborhoods.

Let $C\subseteq X$ be the image of $\{{\theta }_0\}\times S^1$. At a point of $X\setminus C$, a neighborhood is a disk, whose link is a circle. At a point of $C\setminus \{p\}$, three half-disks meet along their diameters, and the link is a Theta graph, with first homology ${\mathbb{Z}}^2$. At $p$, the two folds created by collapsing the horizontal circle give a dumbbell graph as link: two circles joined by an interval. Its first homology is again ${\mathbb{Z}}^2$. The local homology from a link therefore givesBecause local homology is invariant under a homeomorphism, every self-homeomorphism of $X$ preserves the rank-two locus $C$.

It must also preserve $p$. Indeed, $p$ is the unique point of $C$ whose sufficiently small local link is a dumbbell graph; every other point of $C$ has a theta-graph link. The connecting edge of a dumbbell graph is a bridge in a graph, whereas no edge of a theta graph is a bridge, so these local topological types are distinct.

At a point of the open two-cell, the local link is a circle, so local homology in degree two has rank one. At a point of $Z$, a neighborhood consists of $n$ half-discs sharing their boundary interval; its link is the Theta graph ${\Theta }_n$. The local homology from a link calculation givesWhen $n>2$, this local rank characterizes the points of $Z$. Since local homology is invariant under a homeomorphism, every homeomorphism satisfies $f(Z)=Z$, and in particular $f(Z)\subset Z$.

For $n=2$, the two pages form an ordinary disc neighborhood, and $Y$ is ${\mathbb{RP}}^2$. The distinguished circle $Z$ is a projective line, but a projective linear homeomorphism can carry it to a different projective line. Thus the conclusion does not hold.

Yes. Let $b$ be the vertex of $({\Delta }^3{)}^{\prime }$ corresponding to the whole tetrahedron. Its link in $M$ is the 1-skeleton of the barycentric subdivision of $\partial {\Delta }^3$. This connected graph has one vertex for each of the 14 nonempty proper faces and 36 edges, so

No other point has this local homology rank. At the other vertices, corresponding respectively to faces of size $1,2,3$, the first Betti numbers of the links are $6,4,6$. A point in the interior of an edge is incident with at most six triangles, so its local $H_2$ has rank at most five; a point in a triangle interior has local $H_2\cong \mathbb{Z}$.

A homeomorphism preserves local homology, so it must send the unique point with local $H_2$ of rank $23$ to itself. Therefore every self-homeomorphism fixes $b$, which is the fixed barycentre of the barycentric tetrahedral two-skeleton.