[proofplan] The proof proceeds in three movements. First, we package the degree as the unique scalar by which $f^*$ acts on top de Rham cohomology $H^n_{\mathrm{dR}}(S^n) \cong \mathbb{R}$; this delivers well-definedness immediately. Second, smooth homotopy invariance follows by constructing — via fiber integration on the cylinder $S^n \times [0,1]$ — a primitive $\eta$ with $g^*\omega_0 - f^*\omega_0 = d\eta$, after which [Stokes' theorem](/theorems/1530) on the closed manifold $S^n$ collapses the difference. Third, given a regular value $y$, the [inverse function theorem](/page/Inverse%20Function%20Theorem) decomposes a small neighbourhood $B$ of $y$ as the diffeomorphic image of disjoint coordinate balls around the (finitely many) preimage points; choosing $\omega_0$ concentrated in $B$ reduces $\int_{S^n} f^*\omega_0$ to a sum of orientation-signed unit contributions via the change-of-variables formula. Integrality is a corollary, since regular values exist by Sard's theorem and the signed count is manifestly an integer. [/proofplan]

[step:Recast the degree as the scalar action on top de Rham cohomology]The closed oriented manifold $S^n$ has top de Rham cohomology of dimension one, with the integration pairing \begin{align*} \int_{S^n} : H^n_{\mathrm{dR}}(S^n) &\to \mathbb{R}, \\ [\alpha] &\mapsto \int_{S^n} \alpha \end{align*} furnishing an isomorphism by the [Top Cohomology of Orientable Manifolds](/theorems/1531) result. We verify the hypotheses: $S^n$ is a smooth, compact, connected, oriented $n$-manifold without boundary, all of which hold for the standard sphere. Let $\omega_0, \omega_0' \in \Omega^n(S^n)$ be two smooth top forms with $\int_{S^n} \omega_0 = \int_{S^n} \omega_0' = 1$. Their difference satisfies $\int_{S^n}(\omega_0 - \omega_0') = 0$, hence $[\omega_0 - \omega_0'] = 0$ in $H^n_{\mathrm{dR}}(S^n)$ by injectivity of the integration isomorphism. Therefore there exists $\beta \in \Omega^{n-1}(S^n)$ with \begin{align*} \omega_0 - \omega_0' = d\beta. \end{align*} Pulling back under $f$ (which commutes with $d$) gives $f^*\omega_0 - f^*\omega_0' = d(f^*\beta)$, and applying [Stokes' Theorem](/theorems/1530) to the closed manifold $S^n$ (no boundary, so the boundary integral vanishes): \begin{align*} \int_{S^n} f^*\omega_0 - \int_{S^n} f^*\omega_0' = \int_{S^n} d(f^*\beta) = \int_{\partial S^n} f^*\beta = 0. \end{align*} Thus $\deg(f)$ is independent of the unit-integral representative. Equivalently: $f^*$ descends to an endomorphism of $H^n_{\mathrm{dR}}(S^n) \cong \mathbb{R}$, hence acts by a unique scalar; that scalar equals $\deg(f)$.[/step]

[guided]The aim of this step is to verify that the formula $\deg(f) = \int_{S^n} f^*\omega_0$ defines a well-posed real number — i.e., does not depend on which unit-integral $\omega_0$ we picked — and to recast it as multiplication by a scalar on a one-dimensional [vector space](/page/Vector%20Space), which will be the right vantage point for the rest of the proof. The structural fact at work is the [Top Cohomology of Orientable Manifolds](/theorems/1531) theorem, which says that for a closed, connected, oriented smooth $n$-manifold $M$ the integration map \begin{align*} \int_M : H^n_{\mathrm{dR}}(M) \to \mathbb{R} \end{align*} is an isomorphism. We check each hypothesis for $M = S^n$: smoothness (standard atlas), compactness (closed bounded subset of $\mathbb{R}^{n+1}$), connectedness ($S^n$ is path-connected for $n \ge 1$), orientability (induced from $\mathbb{R}^{n+1}$ via the outward normal), and absence of boundary ($\partial S^n = \emptyset$). With this in hand, take two candidate unit-integral forms $\omega_0, \omega_0'$. Their difference has integral zero, so its cohomology class is zero — and "cohomology class zero" means *exact*: there is $\beta \in \Omega^{n-1}(S^n)$ with $\omega_0 - \omega_0' = d\beta$. Now we wish to show the pullbacks integrate to the same number. The pullback operator $f^*$ is a chain map ($f^* \circ d = d \circ f^*$), so \begin{align*} f^*\omega_0 - f^*\omega_0' = f^*(d\beta) = d(f^*\beta). \end{align*} The right-hand side is exact on $S^n$, and exact top forms on a closed manifold integrate to zero — which is precisely [Stokes' Theorem](/theorems/1530) applied to a manifold without boundary: \begin{align*} \int_{S^n} d(f^*\beta) = \int_{\partial S^n} f^*\beta = 0, \end{align*} since $\partial S^n = \emptyset$. Subtracting gives $\int_{S^n} f^*\omega_0 = \int_{S^n} f^*\omega_0'$. The conceptual upshot: the map $f^*: H^n_{\mathrm{dR}}(S^n) \to H^n_{\mathrm{dR}}(S^n)$ is a linear endomorphism of a one-dimensional [vector space](/page/Vector%20Space), so it must be multiplication by a unique scalar, and that scalar is by definition $\deg(f)$. This identification will be the platform for both homotopy invariance (Step 2) and the localisation argument (Step 3).[/guided]

[step:Construct a primitive for $g^*\omega_0 - f^*\omega_0$ via fiber integration on the cylinder]Let $H \in C^\infty(S^n \times [0,1], S^n)$ be a smooth homotopy with $H(\cdot, 0) = f$, $H(\cdot, 1) = g$. Define the inclusion maps \begin{align*} i_t: S^n &\to S^n \times [0,1] \\ x &\mapsto (x, t) \end{align*} for $t \in [0,1]$, so that $f = H \circ i_0$ and $g = H \circ i_1$. We construct a fiber-integration operator $K: \Omega^k(S^n \times [0,1]) \to \Omega^{k-1}(S^n)$ as follows. Any $\alpha \in \Omega^k(S^n \times [0,1])$ decomposes uniquely as \begin{align*} \alpha = \alpha^{(0)} + dt \wedge \alpha^{(1)}, \end{align*} where $\alpha^{(0)} \in \Omega^k(S^n \times [0,1])$ and $\alpha^{(1)} \in \Omega^{k-1}(S^n \times [0,1])$ are characterised by the requirement that they contain no $dt$ factor (in any local product chart). For each $x \in S^n$, the form $\alpha^{(1)}(x, \cdot)$ may be regarded as a smooth one-parameter family of $(k-1)$-forms on $S^n$. Define \begin{align*} (K\alpha)_x := \int_0^1 \alpha^{(1)}(x, t) \, d\mathcal{L}^1(t), \end{align*} the integration being carried out coefficient-by-coefficient in any local frame on $S^n$. A direct computation (separating the $S^n$-derivatives from the $t$-derivative in $d_{S^n \times [0,1]} = d_{S^n} + dt \wedge \partial_t$) yields the **chain homotopy identity**: \begin{align*} i_1^* \alpha - i_0^* \alpha = d(K\alpha) + K(d\alpha) \qquad \text{for all } \alpha \in \Omega^k(S^n \times [0,1]). \end{align*} Apply this with $\alpha := H^*\omega_0 \in \Omega^n(S^n \times [0,1])$. Since $\omega_0$ is a top form on $S^n$, $d\omega_0 = 0$, hence $d(H^*\omega_0) = H^*(d\omega_0) = 0$. The chain homotopy identity therefore gives \begin{align*} g^*\omega_0 - f^*\omega_0 = i_1^*(H^*\omega_0) - i_0^*(H^*\omega_0) = d\bigl(K(H^*\omega_0)\bigr). \end{align*} Set $\eta := K(H^*\omega_0) \in \Omega^{n-1}(S^n)$; then $g^*\omega_0 - f^*\omega_0 = d\eta$.[/step]

[guided]We want to compare the two integrals $\int_{S^n} f^*\omega_0$ and $\int_{S^n} g^*\omega_0$. The cleanest way to do this is to display $g^*\omega_0 - f^*\omega_0$ as an exact form — once we have that, [Stokes' Theorem](/theorems/1530) on the closed manifold $S^n$ will collapse the integral of the difference to zero. The mechanism that produces such a primitive is the **fiber integration** operator on the cylinder $S^n \times [0,1]$, which we now build by hand. First, package the homotopy. We are given $H: S^n \times [0,1] \to S^n$ smooth with $H(\cdot, 0) = f$ and $H(\cdot, 1) = g$. Introducing the slice inclusions $i_t(x) := (x, t)$, the data takes the symmetric form $f = H \circ i_0$, $g = H \circ i_1$, so pullback factors as $f^* = i_0^* \circ H^*$ and $g^* = i_1^* \circ H^*$. The task therefore reduces to comparing $i_0^*$ and $i_1^*$ on forms living on the cylinder. Next, decompose forms on the cylinder. The cylinder $S^n \times [0,1]$ carries the canonical 1-form $dt$, and any smooth $k$-form $\alpha$ splits uniquely as \begin{align*} \alpha = \alpha^{(0)} + dt \wedge \alpha^{(1)} \end{align*} where $\alpha^{(0)}$ has no $dt$ component and $\alpha^{(1)}$ is one degree lower. Geometrically $\alpha^{(0)}$ encodes the "spatial" part and $\alpha^{(1)}$ encodes the "$t$-direction" part. Define the operator $K$ that integrates out the $t$-direction part: \begin{align*} (K\alpha)_x := \int_0^1 \alpha^{(1)}(x, t) \, d\mathcal{L}^1(t). \end{align*} Concretely, in any local product chart on $S^n \times [0,1]$, each coefficient function of $\alpha^{(1)}$ is integrated in $t$ over $[0,1]$. The key identity, which is the algebraic content of de Rham homotopy invariance, is \begin{align*} i_1^* \alpha - i_0^* \alpha = d(K\alpha) + K(d\alpha). \end{align*} To verify it, expand $d = d_{S^n} + dt \wedge \partial_t$ acting on $\alpha = \alpha^{(0)} + dt \wedge \alpha^{(1)}$ and observe that $i_t^*(dt \wedge \alpha^{(1)}) = 0$ (since $i_t^* dt = 0$), so $i_1^* \alpha - i_0^* \alpha$ involves only the $\alpha^{(0)}$ piece; on the other hand, the [fundamental theorem of calculus](/theorems/632) in the $t$-variable contributes $\int_0^1 \partial_t \alpha^{(0)} dt$, which is exactly what the identity records. Now apply this with $\alpha = H^*\omega_0$. Why does this choice cause $K(d\alpha)$ to vanish? Because $\omega_0$ is a top-degree form on $S^n$ (degree $n$), so $d\omega_0 \in \Omega^{n+1}(S^n) = \{0\}$, and pullback commutes with $d$: $d(H^*\omega_0) = H^*(d\omega_0) = 0$. The identity collapses to \begin{align*} i_1^*(H^*\omega_0) - i_0^*(H^*\omega_0) = d\bigl(K(H^*\omega_0)\bigr), \end{align*} i.e., $g^*\omega_0 - f^*\omega_0 = d\eta$ with $\eta := K(H^*\omega_0) \in \Omega^{n-1}(S^n)$. The exact-form expression for the difference is exactly what the next step will integrate.[/guided]

[step:Apply Stokes' theorem to conclude homotopy invariance] By the previous step, $g^*\omega_0 - f^*\omega_0 = d\eta$ on $S^n$. Apply [Stokes' Theorem](/theorems/1530) to the closed oriented $n$-manifold $S^n$ (no boundary): \begin{align*} \int_{S^n} \bigl(g^*\omega_0 - f^*\omega_0\bigr) = \int_{S^n} d\eta = \int_{\partial S^n} \eta = 0, \end{align*} the last equality because $\partial S^n = \emptyset$. Hence \begin{align*} \deg(g) - \deg(f) = \int_{S^n} g^*\omega_0 - \int_{S^n} f^*\omega_0 = 0, \end{align*} proving homotopy invariance (part 2 of the theorem). [/step]

[step:Localise around a regular value via the inverse function theorem]Suppose $y \in S^n$ is a regular value of $f$, so that for every $x \in f^{-1}(y)$ the [linear map](/page/Linear%20Map) $Df_x: T_xS^n \to T_yS^n$ is surjective. Since both tangent spaces are $n$-dimensional, surjectivity equals bijectivity by the [rank-nullity theorem](/theorems/916); thus $Df_x$ is a linear isomorphism, and equivalently the Jacobian matrix $Jf_x \in \mathbb{R}^{n \times n}$ (in any oriented charts around $x$ and $y$) satisfies $\det Jf_x \neq 0$. By the [Inverse Function Theorem](/theorems/51) (applied in coordinate charts — the hypotheses are met since $f$ is smooth and $Df_x$ is invertible at every point of $f^{-1}(y)$), each $x \in f^{-1}(y)$ admits an open neighbourhood $V_x \subset S^n$ such that $f|_{V_x}: V_x \to f(V_x)$ is a diffeomorphism onto an open neighbourhood of $y$. Because $f^{-1}(y)$ is a closed subset of the [compact space](/page/Compact%20Space) $S^n$, it is compact. The argument above shows each preimage point is isolated (a neighbourhood $V_x$ contains no other preimage of $y$, since $f|_{V_x}$ is injective), so $f^{-1}(y)$ is a discrete compact set, hence **finite**: enumerate it as $f^{-1}(y) = \{x_1, \dots, x_N\}$ for some $N \in \mathbb{N}_0$. Shrink the $V_{x_k}$ so that they are pairwise disjoint and so that their images all contain a common open neighbourhood of $y$. Concretely: let $W$ be the intersection $\bigcap_{k=1}^N f(V_{x_k})$, an open neighbourhood of $y$. Choose an open ball $B \subset W$ around $y$ such that $\overline{B}$ is contained in a single oriented chart around $y$ and is small enough that the components $U_k := V_{x_k} \cap f^{-1}(B)$ are pairwise disjoint. Then each \begin{align*} f|_{U_k}: U_k &\to B \end{align*} is a diffeomorphism (the restriction of a diffeomorphism to the preimage of an open subset of the target). We further claim $f^{-1}(B) = \bigsqcup_{k=1}^N U_k$. The inclusion $\bigsqcup U_k \subset f^{-1}(B)$ is by construction. For the reverse, suppose $z \in f^{-1}(B)$ and shrink $B$ if necessary so that this holds: if not, there would be a sequence $z_m \in f^{-1}(B_m)$ with $B_m$ a decreasing sequence of balls around $y$ shrinking to $\{y\}$ such that $z_m \notin \bigcup_k U_k$. By compactness of $S^n$, pass to a subsequence converging to some $z_\infty \in S^n$; by continuity, $f(z_\infty) = y$, so $z_\infty \in f^{-1}(y) = \{x_1, \dots, x_N\}$, contradicting $z_m \notin \bigcup U_k \ni z_\infty$ for all large $m$. (After this shrinking, denote the resulting ball still by $B$ and the disjoint preimage components still by $U_1, \dots, U_N$.)[/step]

[guided]Our overall strategy is to choose the test form $\omega_0$ supported in a tiny ball $B$ around $y$, so that $f^*\omega_0$ is supported in $f^{-1}(B)$. To turn $\int_{S^n} f^*\omega_0$ into a finite signed sum, we need $f^{-1}(B)$ to break up as a disjoint union of small open sets, each mapped diffeomorphically onto $B$ — one piece per preimage point. This step manufactures that decomposition. **Why is $Df_x$ an isomorphism?** Regularity of the value $y$ says $Df_x$ is surjective. Both $T_xS^n$ and $T_yS^n$ are $n$-dimensional real vector spaces, and a surjective [linear map](/page/Linear%20Map) between finite-dimensional spaces of equal dimension is automatically injective (rank-nullity). Hence $Df_x$ is an isomorphism, and in any local chart its Jacobian matrix has nonzero determinant. **Why does the [inverse function theorem](/theorems/51) apply?** The [Inverse Function Theorem](/theorems/51) is a statement about smooth maps between open subsets of Euclidean space whose total derivative at a point is invertible. We apply it after composing with oriented coordinate charts $\varphi$ near $x$ and $\psi$ near $y$: the map $\psi \circ f \circ \varphi^{-1}$ is smooth (since $f$ is smooth and the charts are smooth), and its total derivative at $\varphi(x)$ is the matrix representation of $Df_x$, which is invertible. The IFT delivers a smooth local inverse, which pulls back through the charts to a diffeomorphism $V_x \to f(V_x)$. **Why is $f^{-1}(y)$ finite?** It is closed (preimage of the [closed set](/page/Closed%20Set) $\{y\}$ under the continuous map $f$) in the [compact space](/page/Compact%20Space) $S^n$, hence itself compact. Each preimage point is isolated by the [inverse function theorem](/page/Inverse%20Function%20Theorem) (no two preimage points can lie in the same $V_x$, since $f|_{V_x}$ is injective). A discrete compact subset of a [Hausdorff space](/page/Hausdorff%20Space) is finite, so $f^{-1}(y) = \{x_1, \dots, x_N\}$. **Why can we make the preimages of one ball disjoint?** This is a standard compactness argument. After choosing disjoint $V_{x_k}$, the obstruction would be a sequence of points $z_m$ with $f(z_m) \to y$ but $z_m$ outside every $V_{x_k}$. By compactness of $S^n$ such a sequence has a limit point $z_\infty$ with $f(z_\infty) = y$, but then $z_\infty$ is one of the $x_k$, contradicting $z_m \notin V_{x_k}$ for large $m$. Hence for a sufficiently small ball $B$ around $y$, $f^{-1}(B) \subset \bigcup_k V_{x_k}$, and intersecting with each $V_{x_k}$ yields the desired disjoint decomposition. The end result is a "downstairs" ball $B$ around $y$ whose preimage $f^{-1}(B)$ consists of $N$ disjoint "upstairs" open sets $U_1, \dots, U_N$, each diffeomorphic via $f$ to $B$.[/guided]

[step:Compute the local contribution on each component by change of variables]Choose a unit-integral top form $\omega_0 \in \Omega^n(S^n)$ that is **supported in** $B$ — this is possible because $B$ is a nonempty open subset of $S^n$ and the standard volume form can be multiplied by a smooth bump that is positive in $B$ and zero outside, then rescaled to total mass $1$. By Step 1 (independence of representative), this choice does not change $\deg(f)$. Since $f^*\omega_0$ is supported in $f^{-1}(\operatorname{supp} \omega_0) \subset f^{-1}(B) = \bigsqcup_{k=1}^N U_k$, \begin{align*} \deg(f) = \int_{S^n} f^*\omega_0 = \sum_{k=1}^N \int_{U_k} f^*\omega_0. \end{align*} Fix $k$. The restriction $f|_{U_k}: U_k \to B$ is a diffeomorphism. Choose oriented charts $\varphi: U_k \to \widetilde U_k \subset \mathbb{R}^n$ (induced from the orientation of $S^n$ at $x_k$) and $\psi: B \to \widetilde B \subset \mathbb{R}^n$ (induced at $y$); the transition map $F_k := \psi \circ f \circ \varphi^{-1}: \widetilde U_k \to \widetilde B$ is a diffeomorphism between open subsets of $\mathbb{R}^n$. Write $\omega_0|_B = u \cdot dy_1 \wedge \cdots \wedge dy_n$ pulled back through $\psi$ — more precisely, $(\psi^{-1})^*\omega_0 = \tilde u(y) \, dy_1 \wedge \cdots \wedge dy_n$ on $\widetilde B$, where $\tilde u: \widetilde B \to \mathbb{R}$ is smooth with $\int_{\widetilde B} \tilde u \, d\mathcal{L}^n(y) = \int_B \omega_0 = \int_{S^n} \omega_0 = 1$. Pulling back by $F_k$: \begin{align*} F_k^*\bigl(\tilde u(y) \, dy_1 \wedge \cdots \wedge dy_n\bigr) = \tilde u(F_k(x)) \, \det(JF_{k,x}) \, dx_1 \wedge \cdots \wedge dx_n, \end{align*} where $JF_{k,x} \in \mathbb{R}^{n \times n}$ is the Jacobian matrix of $F_k$ at $x$. Integrating against the chart orientation, \begin{align*} \int_{U_k} f^*\omega_0 &= \int_{\widetilde U_k} \tilde u(F_k(x)) \, \det(JF_{k,x}) \, d\mathcal{L}^n(x). \end{align*} Since $F_k$ is a diffeomorphism and $\widetilde U_k$ is connected (it is the image of a connected $U_k$ under a chart; choose $U_k$ connected from the outset by replacing it with the connected component of $f^{-1}(B)$ containing $x_k$, which is again open and diffeomorphic to $B$ via $f$), the continuous nonvanishing function $\det(JF_{k,x})$ has constant sign on $\widetilde U_k$. Denote this sign by $\varepsilon_k := \operatorname{sgn}(\det Jf_{x_k}) \in \{+1, -1\}$ — it agrees with the sign at the single point $x_k$ because the charts are oriented and the transition functions between oriented charts have positive Jacobian determinant. Apply the [Change of Variables (general)](/theorems/22) theorem to the diffeomorphism $F_k: \widetilde U_k \to \widetilde B$ between open subsets of $\mathbb{R}^n$, with integrand $\tilde u$ on $\widetilde B$. The hypotheses are met: $F_k$ is a $C^1$-diffeomorphism (in fact smooth), the integrand $\tilde u$ is Borel measurable (smooth, in particular continuous), and the change-of-variables identity $\int_{\widetilde B} \tilde u \, d\mathcal{L}^n = \int_{\widetilde U_k} \tilde u \circ F_k \cdot |\det JF_k| \, d\mathcal{L}^n$ holds. Combining with the sign $\varepsilon_k = \det(JF_{k,x})/|\det(JF_{k,x})|$ (constant on $\widetilde U_k$): \begin{align*} \int_{U_k} f^*\omega_0 &= \int_{\widetilde U_k} \tilde u(F_k(x)) \, \det(JF_{k,x}) \, d\mathcal{L}^n(x) \\ &= \varepsilon_k \int_{\widetilde U_k} \tilde u(F_k(x)) \, |\det(JF_{k,x})| \, d\mathcal{L}^n(x) \\ &= \varepsilon_k \int_{\widetilde B} \tilde u(y) \, d\mathcal{L}^n(y) \\ &= \varepsilon_k \cdot 1 = \operatorname{sgn}(\det Jf_{x_k}). \end{align*}[/step]

[guided]The arithmetic of this step is the heart of the theorem: each preimage point contributes exactly $\pm 1$ to the degree, with the sign determined by whether $Df_{x_k}$ preserves or reverses orientation. **Why can we choose $\omega_0$ supported in $B$?** Because by Step 1 we proved $\deg(f)$ is independent of the choice of unit-integral top form. So we are free to pick any representative — and the right one for the computation is a bump form concentrated in the small ball $B$. Concretely, take the standard Riemannian volume form on $S^n$, multiply by a smooth bump $\rho \in C_c^\infty(B)$ with $\rho \geq 0$ and $\rho > 0$ somewhere, and rescale so the total integral is $1$. The resulting $\omega_0$ has support inside $B$. **Why does $f^*\omega_0$ live inside $\bigsqcup U_k$?** A pullback's support is contained in the preimage of the original support: $\operatorname{supp}(f^*\omega_0) \subset f^{-1}(\operatorname{supp} \omega_0) \subset f^{-1}(B)$. By Step 3 the right-hand side is exactly $\bigsqcup_{k=1}^N U_k$, so the integral over $S^n$ collapses to a sum of $N$ integrals, one per preimage point. **How do charts enter?** Each $U_k$ is diffeomorphic to $B$ via $f|_{U_k}$, but to apply the Euclidean change-of-variables theorem we need to land in $\mathbb{R}^n$. Choose oriented charts $\varphi$ near $x_k$ and $\psi$ near $y$ — orientation of the chart means that $\varphi_* (\text{orientation of } S^n \text{ at } x_k)$ equals the standard orientation of $\mathbb{R}^n$, and likewise for $\psi$. The composition $F_k = \psi \circ f \circ \varphi^{-1}$ is then a smooth diffeomorphism between open subsets of $\mathbb{R}^n$, and the orientation-preserving / orientation-reversing nature of $Df_{x_k}$ translates directly to the sign of $\det JF_{k,x_k}$. **Why does the Jacobian determinant have constant sign?** On a connected [open set](/page/Open%20Set), a continuous nonvanishing real-valued function has constant sign ([intermediate value theorem](/theorems/180)). The Jacobian $\det JF_{k,x}$ is continuous in $x$ and nonzero (because $F_k$ is a diffeomorphism, so its derivative is invertible everywhere). By shrinking $U_k$ to be connected, we ensure $\det JF_{k,x}$ has a single sign $\varepsilon_k$ throughout, which equals its value at the central point $x_k$. **How does the algebra of the change of variables work?** The pullback of $dy_1 \wedge \cdots \wedge dy_n$ by $F_k$ is the differential-forms version of the Jacobian formula: \begin{align*} F_k^*(dy_1 \wedge \cdots \wedge dy_n) = \det(JF_{k,x}) \, dx_1 \wedge \cdots \wedge dx_n. \end{align*} The crucial sign sits in $\det(JF_{k,x})$, **not** $|\det(JF_{k,x})|$ — orientation matters for differential forms. Integrating an oriented form over an oriented chart is what the chart-based definition $\int_{U_k} \alpha := \int_{\widetilde U_k}$ (coefficient) $d\mathcal{L}^n$ encodes. The standard Euclidean [Change of Variables (general)](/theorems/22) theorem, which uses absolute values, then converts $\int_{\widetilde U_k}$ to $\int_{\widetilde B}$ at the cost of a single $\varepsilon_k$ sign: \begin{align*} \int_{\widetilde U_k} \tilde u(F_k(x)) \det(JF_{k,x}) \, d\mathcal{L}^n(x) = \varepsilon_k \int_{\widetilde U_k} \tilde u(F_k(x)) |\det(JF_{k,x})| \, d\mathcal{L}^n(x) = \varepsilon_k \int_{\widetilde B} \tilde u(y) \, d\mathcal{L}^n(y). \end{align*} Since $\tilde u$ was chosen to integrate to $1$, the contribution is $\varepsilon_k = \operatorname{sgn}(\det Jf_{x_k})$.[/guided]

[step:Sum the local contributions and deduce the signed-count formula] Summing the local contributions from the previous step: \begin{align*} \deg(f) = \sum_{k=1}^N \int_{U_k} f^*\omega_0 = \sum_{k=1}^N \operatorname{sgn}(\det Jf_{x_k}) = \sum_{x \in f^{-1}(y)} \operatorname{sgn}(\det Jf_x), \end{align*} which is part 3 of the theorem. [/step]

[step:Conclude integrality of the degree] By Sard's theorem (citing a result not yet in the wiki: **Sard's Theorem**, which states that for a smooth map $f: M \to N$ between smooth manifolds of equal dimension, the set of critical values has Lebesgue measure zero in $N$), the set of regular values of $f: S^n \to S^n$ has full measure in $S^n$ and is in particular nonempty. Pick any regular value $y_0 \in S^n$. By the previous step, \begin{align*} \deg(f) = \sum_{x \in f^{-1}(y_0)} \operatorname{sgn}(\det Jf_x). \end{align*} The right-hand side is a finite sum of $\pm 1$, hence an integer. This completes part 1 of the theorem and concludes the proof. [/step]