Degree of a continuous mapping ($\mathrm{deg}f$)

For a map $f:S^m\to S^m$, the degree is the integer by which $f_{\ast }$ multiplies a chosen generator of $H_m(S^m;\mathbb{Z})$. It is invariant under homotopy.

For a map $f:M\to N$ between closed connected $n$-manifolds, the mod-two degree is defined by $f_{\ast }[M{]}_2={\mathrm{deg}}_2(f)[N{]}_2$. Every manifold has a mod-two orientation class, so no orientability assumption is needed. It is invariant under homotopy. A nonsurjective map has mod-two degree zero: mapping the fundamental classes into the local relative homology at a missed target point makes the source contribution zero but leaves the target local generator nonzero.

Suppose a map $S^m\to M$ of closed connected oriented manifolds has nonzero mapping degree. A product decomposition with $0<k<m$ would have oriented closed factors: an orientation-reversing loop in either factor, with the other coordinate fixed, would reverse the product orientation. Pull back their top orientation classes to obtain $\alpha \in H^k(M;\mathbb{Z})$ and $\beta \in H^{m-k}(M;\mathbb{Z})$ with $\alpha ⌣\beta$ a generator of $H^m(M;\mathbb{Z})$. Both pullbacks to the sphere vanish in their intermediate degrees, contradicting the nonzero-degree pullback of their product.

Define mapping degree for $S^0$ using its rank-one reduced homology group ${\tilde{H}}_0(S^0;\mathbb{Z})$. Its generator is the difference of its two points. The four maps of that two-point set consist of the identity, the interchange, and two constant maps; their degree multipliers are $1,-1,0,0$. Thus sphere maps of arbitrary integer degree require positive sphere dimension.

For a map of oriented disk pairs, the multiplier on top relative homology is its relative mapping degree. Naturality of the connecting homomorphism identifies it with the mapping degree of the boundary map.

For a smooth map of disk pairs whose graph of a function is transverse to the zero slice, orient the graph by its domain. With the tangent space of the zero slice ordered before that of the graph, the local smooth intersection number at a zero is $\mathrm{sgn}\det DF$. Summing these signs gives the relative mapping degree and hence the boundary mapping degree. Reversing the order of these two $m$-dimensional tangent spaces changes the sign by $(-1{)}^m$.

Collapsing the boundary identifies the top relative homology of a disk pair with the top reduced homology of its quotient sphere. Together with the boundary connecting homomorphism, this shows that the induced quotient map and the boundary map have the same mapping degree, using the corresponding orientations.

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.

In dimension two, $z\mapsto z^k$ for positive $k$ has local degree $k$ at zero, and $z\mapsto {\overline{z}}^{|k|}$ has degree $k$ for negative $k$. The proper map $z\mapsto |z{|}^2\in \mathbb{C}$ has isolated zero fiber and local degree zero, since its image misses every nearby nonreal target. These proper maps extend to the two-sphere. Taking their product with the identity of ${\mathbb{R}}^{n-2}$ realizes the same local degrees in every dimension $n\ge 2$. In dimension one an isolated local degree is restricted to $0,1,-1$.

For a continuous map $f:M\to N$ of closed connected oriented $d$-dimensional manifolds, suppose $f^{-1}(y)=\{x_1,\dots ,x_s\}$ is finite. ThenBy excision, the degree-$d$ relative homology group $H_d(M,M\setminus f^{-1}(y);\mathbb{Z})$ is the direct sum of the local groups at $x_j$. The fundamental class maps to the tuple of local orientation generators. Mapping this tuple to $H_d(N,N\setminus \{y\};\mathbb{Z})$ adds their local degrees. Naturality identifies the result with the image of $(\mathrm{deg}f)[N]$. If the fiber is empty, the same argument gives degree zero.

If $f:S^m\to S^m$, then its suspension $\Sigma f:S^{m+1}\to S^{m+1}$ has the same degree. This follows from naturality of the suspension isomorphism in reduced homology.

Every integer $m$ is the mapping degree of a continuous map $S^n\to S^n$ in every dimension $n\ge 1$. Start with the circle map $z\mapsto z^m$, whose argument winds $m$ times, and use degree under suspension repeatedly. Negative $m$ reverses the winding, and $m=0$ is constant before suspending. The degrees of maps of the zero-sphere are a genuine exceptional case.

For a map $f:M\to N$ between closed connected oriented $d$-manifolds, its degree is the integer determined by $f_{\ast }[M]=(\mathrm{deg}f)[N]$ in top-dimensional homology.

Let $M,N$ be closed connected oriented manifolds of the same dimension, and suppose $a^r$ generates the top integral cohomology of $N$. If every class of degree $|a|$ on $M$ has zero rth cup power, every continuous map $f:M\to N$ has mapping degree zero. Indeed, $f^{\ast }(a^r)=(f^{\ast }a{)}^r=0$, and evaluation on the fundamental class gives $\mathrm{deg}f=0$. For $S^2\times S^4\to {\mathbb{CP}}^3$, the degree-two class is a multiple of the sphere class $u$, whose square vanishes.

For $k>0$, an epimorphism ${\pi }_1({\Sigma }_2)\to \mathbb{Z}/k$ gives a connected $k$-sheeted cover. Its Euler characteristic is $-2k$, hence its genus is $k+1$, and the lifted orientation gives degree $k$. For negative $k$ compose the $|k|$-sheeted cover with an orientation-reversing homeomorphism of the target. For $k=0$ take a constant map from the torus.

Choose an oriented embedded closed disc in a closed oriented positive-dimensional manifold and collapse its complement to a point. A target point away from the collapsed value has one preimage with local degree $+1$, proving global degree one by the degree as a sum of local degrees.

On the unit two-sphere this formula maps onto the torus: every torus point has arguments corresponding to $x,y\in [-1/2,1/2]$, and $x^2+y^2\le 1/2$ leaves a real $z$ on the sphere. Scaling both arguments continuously to zero is a null-homotopy, so the degree is zero. Surjectivity alone does not force nonzero degree.

For maps of oriented connected closed manifolds of the same dimension, apply functoriality of homology to $f_{\ast }[M]=(\mathrm{deg}f)[N]$ and $g_{\ast }[N]=(\mathrm{deg}g)[L]$. The composite coefficient is their product. This recovers the multiplication of signed sheet counts for compositions of finite covering maps.

For maps between oriented connected closed manifolds of the same dimension, a homotopy induces equal maps on top-dimensional homology, hence equal coefficients on the fundamental class. For smooth maps and a smooth homotopy, the degree by integration of a pullback volume form gives another proof: the Generalized Stokes theorem on the homotopy cylinder makes the difference of endpoint pullback integrals zero because the target top form is closed. This is an obstruction to extending a degree-one sphere identity over a ball.

Although topological degree classifies maps $S^n\to S^n$ up to homotopy, it is not a complete invariant for other manifolds. On the torus, the identity and the integer shear $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but induce distinct maps on $H_1(T^2;\mathbb{Z})$ and cannot be homotopic.

For a smooth map $f:M\to N$ between oriented connected closed manifolds of the same dimension, a normalized volume form on the target gives $\mathrm{deg}f={\int }_M{f}^{\ast }\omega$. The formula agrees with the degree as a sum of local degrees. A change of normalized top-form adds an exact form by top-dimensional de Rham cohomology, whose pullback has zero integral by Stokes theorem.

For a smooth map between oriented unit spheres, integrate the pullback of the target area form. Localizing a normalized top-degree form near a regular value, and using Stokes theorem to discard exact-form differences, identifies this integral with the degree as a sum of local degrees.

A degree of a map between oriented manifolds equal to one makes the pullback on degree-one cohomology injective by the Poincare duality pairing. Thus a degree-one map of closed oriented surfaces requires $2h\le 2g$. Conversely, collapse the extra handles in ${\Sigma }_h\#{\Sigma }_{g-h}$ onto a point. The retained oriented disc shows that this map has degree one.

If $S^n\to N$ has prime degree $p$ and $N$ is a closed connected oriented $n$-manifold, every integral homology group of $N$ in degrees strictly between zero and $n$ is a finite $p$-primary abelian group. Consequently one power of $p$ annihilates all these groups. The proof combines Poincare duality over each field ${\mathbb{F}}_q$, $q\ne p$, with the universal coefficient theorem for homology and finite generation.

For $n\ge 2$, a self-map of $S^n$ factoring through Real projective space ${\mathbb{RP}}^n$ has degree zero when $n$ is even, and can have exactly the even degrees when $n$ is odd. In the odd case the first map lifts through the double covering and the covering has degree two. The dimension-one exception allows every integer, since ${\mathbb{RP}}^1\cong S^1$.

The antipodal map $x\mapsto -x$ on $S^m$ has degree $(-1{)}^{m+1}$.

If a continuous map $f:S^d\to S^d$ has no fixed point, the formula $H_t(x)=((1-t)f(x)-tx)/\Vert (1-t)f(x)-tx\Vert$ is a homotopy from $f$ to the antipodal map. The denominator could vanish only at $t=1/2$ and $f(x)=x$, which is excluded. The homotopy invariance of mapping degree therefore gives $\mathrm{deg}f=(-1{)}^{d+1}$.

An odd map between spheres is a continuous map $f:S^n\to S^m$ satisfying $f(-x)=-f(x)$. Thus it is equivariant for the two antipodal maps. It descends to a continuous map of Real projective spaces.

Suppose a continuous map $g:S^n\times S^n\to S^n$ changes sign on negating either input. Its quotient $\bar{g}:{\mathbb{RP}}^n\times {\mathbb{RP}}^n\to {\mathbb{RP}}^n$ pulls back the real tautological line bundle to the tensor product of vector bundles $p_1^{\ast }{\gamma }_n\otimes p_2^{\ast }{\gamma }_n$. The fiber map sends $(sx)\otimes (ty)$ to $stg(x,y)$. The First Stiefel–Whitney class of a tensor product of real line bundles consequently gives ${\bar{g}}^{\ast }a=a_1+a_2$. The Künneth theorem and the mod-two cohomology ring of real projective space implyEvery interior binomial coefficient in row $n+1$ must be even. By binomial coefficients with even interior terms, $n=2^k-1$ for some $k\ge 0$. This is a necessary condition, not an assertion of existence in every such dimension.

For an odd map between spheres $f:S^n\to S^m$ and its quotient $\bar{f}$, there is an isomorphism ${\gamma }_n\cong {\bar{f}}^{\ast }{\gamma }_m$ of real tautological line bundles. With $x$ a unit vector, send $tx$ to $tf(x)$; replacing $x$ by $-x$ and $t$ by $-t$ gives the same vector. This proves both linearity on fibers and well-definedness. Therefore ${\bar{f}}^{\ast }{w}_1({\gamma }_m)=w_1({\gamma }_n)$. The mod-two cohomology ring of real projective space then implies $n\le m$, since a vanishing $(m+1)$st power must pull back to a vanishing power.

Let a finite group $G$ act smoothly on a smooth manifold, and suppose that a $G$-invariant differential form $\omega$ is an exact differential form, say $\omega =d\eta$. Averaging gives the invariant primitiveIf the action is free, invariant forms descend uniquely through the resulting covering map.

For the double covering $\pi :S^{2n}\to {\mathbb{RP}}^{2n}$ and the antipodal map $a$, every pulled-back top form satisfies $a^{\ast }{\pi }^{\ast }\omega ={\pi }^{\ast }\omega$. Since $\mathrm{deg}a=-1$, its integral over the sphere is its own negative and hence vanishes. It is therefore exact on the sphere. Averaging a primitive under $a$ and descending it proves that $\omega$ is exact on real projective space.

A Euclidean homeomorphism is a proper map and extends to its one-point compactification $S^n$. Its degree is the degree of a continuous mapping of this extension, and belongs to $\{1,-1\}$. A linear isomorphism has degree equal to the sign of its determinant.

The factor swap ${\mathbb{R}}^m\times {\mathbb{R}}^n\to {\mathbb{R}}^n\times {\mathbb{R}}^m$ has degree $(-1{)}^{mn}$. Moving an oriented basis of the first factor past one of the second requires $mn$ transpositions.