The de Rham Theorem and Degree — Examples

At $t\in [0,1]$, the pullback definition gives $({\sigma }^{\ast }\omega {)}_t({\mathrm{\partial }}_t)={\omega }_{\sigma (t)}(d{\sigma }_t({\mathrm{\partial }}_t))={\omega }_{\sigma (t)}({\sigma }^{\mathrm{\prime }}(t))\mathrm{.}$ Therefore ${\sigma }^{\ast }\omega =f(t)dt$ with $f(t)={\omega }_{\sigma (t)}({\sigma }^{\mathrm{\prime }}(t))$.

Integrating this coefficient in the positive orientation of $[0,1]$ and using [F1] gives the displayed formula. If $\sigma$ is constant then ${\sigma }^{\mathrm{\prime }}=0$ and both sides are zero; if $\omega =0$ the same holds. Both parameter endpoints are included in the smooth-simplex convention and do not change the Riemann integral. The formula uses a single supplied path and no choice principle.

The oriented boundary is $\mathrm{\partial }[v_0{v}_1{v}_2]=[v_1{v}_2]-[v_0{v}_2]+[v_0{v}_1]\mathrm{.}$ Applying [F1] and linearity of chain integration gives exactly the first displayed formula, including its middle minus sign.

For the standard triangle with $v_0=(0,0)$, $v_1=(1,0)$ and $v_2=(0,1)$, one has $d(xdy)=dx\wedge dy$ and therefore ${\int }_{\sigma }d\eta ={\int }_0^1{\int }_0^{1-x}dydx={\int }_0^1(1-x)dx=\frac{1}{2}\mathrm{.}$ On the first boundary edge use $(x,y)=(1-t,t)$, $0\le t\le 1$, so $xdy=(1-t)dt$ and its integral is $1\mathrm{/}2$. On $[v_0{v}_2]$ one has $x=0$, and on $[v_0{v}_1]$ one has $dy=0$, so the other two edge integrals vanish. Thus the signed boundary total is also $1\mathrm{/}2$.

Reversing the simplex orientation reverses both sides and all three induced edge signs. A zero form or degenerate simplex gives zero through [F1]; there is no omitted boundary endpoint because each oriented edge includes both of its vertices. The calculation is finite and choice-free.

Differentiation using [F2] gives $x^{\mathrm{\prime }}(t)=-2\pi \sin (2\pi t)$ and $y^{\mathrm{\prime }}(t)=2\pi \cos (2\pi t)$. Hence ${\gamma }^{\ast }(-ydx+xdy)=(-y(t)x^{\mathrm{\prime }}(t)+x(t)y^{\mathrm{\prime }}(t))dt=2\pi ({\sin }^2(2\pi t)+{\cos }^2(2\pi t))dt=2\pi dt,$ and therefore ${\gamma }^{\ast }\alpha =dt$.

Therefore [F1] gives $I_{S^1}^1(\alpha )(\gamma )={\int }_0^{1}dt=1.$ More generally, composing with $P_m(z)=z^m$ gives the lift $t\mapsto mt$ and $P_m^{\ast }\alpha =m\alpha$, so the same calculation yields $I^1(\alpha )(P_m\gamma )=m$, in agreement with $\mathrm{deg}(P_m)=m$ from [F3], including negative $m$ and $m=0$.

Reversing the loop changes the value to $-1$; the constant loop has value zero. Both endpoints of $[0,1]$ map to the same circle point, so the path is a cycle and there is no seam contribution. The calculation uses explicit maps and no choice principle.

The radial homotopy is $H(x,t)=c+t(x-c)$. For a $k$-form $\omega$ with $k\ge 1$, its homotopy operator is explicitly $(L_H\omega {)}_x(v_1,\dots ,v_{k-1})={\int }_0^1{t}^{k-1}{\omega }_{c+t(x-c)}(x-c,v_1,\dots ,v_{k-1})dt\mathrm{.}$ The homotopy identity in [F1] gives $d{L}_H\omega +L_{H}d\omega =\omega -H_0^{\ast }\omega$; since the constant-map pullback $H_0^{\ast }\omega$ is zero in positive degree, every closed positive-degree form is exact. A closed zero-form satisfies $f(x)=f(c)$, so evaluation at $c$ identifies $H_{\mathrm{d}\mathrm{R}}^0(B)$ with $\mathbb{R}$.

On a smooth singular $q$-simplex $\sigma$, triangulate ${\mathrm{\Delta }}^q\times [0,1]$ by the $q+1$ affine prism simplices ${\kappa }_i$ with ordered vertices $(v_0,0),\dots ,(v_i,0),(v_i,1),\dots ,(v_q,1),$ and put $P\sigma ={\sum }_{i=0}^q(-1{)}^{i}H(\sigma \times \mathrm{id}){\kappa }_i$. The cancellations of paired interior faces leave $\mathrm{\partial }P+P\mathrm{\partial }={\mathrm{id}}_{\mathrm{\#}}-(c{)}_{\mathrm{\#}}\mathrm{.}$ Dualizing gives $\mathrm{id}-c^{\ast }=\delta D+D\delta$, not a contraction of the unnormalized complex by itself: positive-dimensional constant simplices remain nonzero. The point complex is the alternating complex with one generator in every degree, whose positive cohomology vanishes, and [F1] uses this calculation together with the displayed homotopy to prove that inclusion of the centre and constant projection induce inverse cohomology maps. Hence $H_{\mathrm{\infty }}^0(B;\mathbb{R})=\mathbb{R}$ and the positive groups vanish.

Integration sends a constant function $a$ to the zero-cochain taking value $a$ at every vertex. Hence it is the identity under the two degree-zero evaluations, and in positive degrees it is the unique map $0\to 0$. If $n=0$, the ball is a point and the unnormalized constant simplices still contract as in [F1]. Zero forms, constant simplices, and both homotopy endpoints are included above. One centre and explicit finite prism sums are used, so no choice principle enters.

Choose $a\in W_0$ and $b\in W_1$. Let $u$ be the increasing arc in $U$ from $a$ to $b$, and $v$ the increasing arc in $V$ from $b$ through the quotient seam to $a$; $z=u+v$ is the positively oriented circle cycle. For the de Rham lift set $\alpha =-{\rho }_{V}c$ on $U$ and $\beta ={\rho }_{U}c$ on $V$. Their difference on the overlap is $\beta -\alpha =({\rho }_U+{\rho }_V)c=c$, and their derivatives glue to the connecting one-form $\zeta$.

Since $\zeta =d\alpha$ on $U$ and $\zeta =d\beta$ on $V$, endpoint evaluation gives ${\int }_z\zeta =\alpha (b)-\alpha (a)+\beta (a)-\beta (b)\mathrm{.}$ At $b\in W_1$, $c(b)=0$, hence $\alpha (b)=\beta (b)=0$; at $a\in W_0$, $\beta (a)-\alpha (a)=c(a)=1$. Therefore ${\int }_z\zeta =1$.

On the singular side use the lift $e_0=(-E_{U}c,0)$ from [F1]. Its differential glues to the connector cocycle $z_0$. On $u:a\to b$, $z_0(u)=\delta (-E_{U}c)(u)=(-c(b))-(-c(a))=1,$ while $z_0(v)=0$ on the $V$ lift. Thus $z_0(z)=1$, exactly the value in step 2.1, and [F1] identifies the two connector classes.

Replacing $c$ by $-c$ or reversing $z$ reverses both answers. The zero cocycle gives zero on both sides; if an overlap component were absent, this particular nonzero witness would not exist. Both endpoints of each arc occur in the displayed coboundary differences, and degenerate simplices contribute zero. Apart from [F2]'s partition existence, every lift, path, sign and evaluation is finite and explicit.

Take $\rho$ from [F1] and set $a={\int }_{-1}^1\rho (t)dt$. The partition at $-1\mathrm{/}2$ and $1\mathrm{/}2$, together with $\rho \ge 0$ everywhere and $\rho =1$ on the central interval, has lower Darboux sum at least $1$; hence [F2] gives $a\ge 1>0$. Put $b=\rho \mathrm{/}a$. Then $b$ is smooth, supported in $(-1,1)$, and ${\int }_{\mathbb{R}}b=1$.

Using the library's zero-based coordinates on $n=\{0,\dots ,n-1\}$, define $\omega (x_0,\dots ,x_{n-1})=({\prod }_{j=0}^{n-1}b(x_j))d{x}^0\wedge \cdots \wedge d{x}^{n-1}\mathrm{.}$ Its coefficient is smooth and its support lies in the closed cube $[-1,1{]}^n$, which is compact by [F3]. Repeated application of [F3] on that cube gives ${\int }_{{\mathbb{R}}^n}\omega ={\prod }_{j=0}^{n-1}({\int }_{-1}^{1}b(t)dt)=1^n=1.$ Thus [F4] sends $[\omega ]$ to $1$.

For $n=1$ the product and Fubini iteration have one factor and recover $b(t)dt$. With the standard convention in dimension zero, the empty product is the value-one function on the positive point and also has integral one, although the displayed construction was stipulated for $n\ge 1$. Replacing any factor by zero makes the integral zero, as linearity predicts. There are no endpoints of the ambient manifold; the bounding-cube faces only delimit a zero extension. One explicitly constructed bump is reused finitely many times, so no choice principle enters.

The ambient linear map $R$ is orthogonal, has determinant $-1$, preserves the unit sphere, and satisfies $R^{-1}=R$. Hence its restriction is a smooth diffeomorphism. If $(v_1,\dots ,v_n)$ is a positive tangent basis at $x$, then [F1] makes $(x,v_1,\dots ,v_n)$ positive in ${\mathbb{R}}^{n+1}$. The target outward-normal frame is $(Rx,R{v}_1,\dots ,R{v}_n)=R(x,v_1,\dots ,v_n),$ which has the opposite ambient orientation because $\det R=-1$. Thus the tangent image basis is negative at $Rx$.

Therefore the sphere reflection reverses orientation, and [F2] gives $\mathrm{deg}(R)=-1$. For $n=1$ this is the ordinary reflection of a circle across an axis. Points on the reflecting equator are fixed but still have negative tangent sign, so fixed points do not create a degenerate exception. The disconnected case $S^0$ is excluded by $n\ge 1$; there are no manifold-boundary endpoints, and the pointwise determinant calculation makes no choices.

Put $r=\mathrm{\mid }m\mathrm{\mid }$. The $r$ classes $x_k=\left[\frac{a+k}{m}\right],0\le k<r,$ all map to $[a]$. They are distinct: if $x_j=x_k$, then $(j-k)\mathrm{/}m\in \mathbb{Z}$, so $r$ divides $j-k$, which is possible in the displayed range only when $j=k$. Conversely, if $P_m([t])=[a]$, then $mt-a\in \mathbb{Z}$; reducing that integer modulo $r$ puts $[t]$ equal to exactly one $x_k$. Thus this is the complete fibre.

In increasing angular lift coordinates at every $x_k$ and at $y$, the map is $u\mapsto mu$ plus a constant, so its derivative is the nonzero scalar $m$. Every $x_k$ is therefore regular and has local sign $\mathrm{sgn}(m)$. Properness from [F3] lets [F1] apply, giving $\mathrm{deg}(P_m)={\sum }_{k=0}^{r-1}\mathrm{sgn}(m)=r\mathrm{sgn}(m)=m,$ agreeing with [F2].

For $m=1$ the fibre is a singleton of sign $+1$; for $m=-1$ it is a singleton of sign $-1$. The excluded $m=0$ map instead has empty fibres away from its constant value and degree zero, as [F2] records. There are no quotient-seam endpoints: all calculations use local lifts. The finite list is explicit and no choice principle is used.

Let $y=[a]\in S^1$. Its fibre is exactly $F^{-1}(y)=\{\left[\frac{a}{2}\right],\left[\frac{a+1}{2}\right]\}\mathrm{.}$ Both displayed classes map to $[a]$, and they are distinct because their difference is $1\mathrm{/}2\mathrm{\notin }\mathbb{Z}$. Conversely, $F([t])=[a]$ means $2t-a=k\in \mathbb{Z}$; according as $k$ is even or odd, $[t]$ is the first or second displayed class. This also shows that changing $a$ by an integer merely permutes the two classes.

Choose quotient arcs about either preimage and $y$, and lift them to increasing real coordinates. On each source arc $F$ has the form $u\mapsto 2u-k$ for an integer $k$, so its derivative is $2>0$. Hence each restriction is an orientation-preserving diffeomorphism onto a sufficiently short target arc; these two restrictions are the two inverse sheets over that arc. Thus every $y$ is regular and both local signs are $+1$.

Since $F$ is proper by [F3], [F2] applies at the arbitrary value $y$ and gives $\mathrm{deg}(F)=(+1)+(+1)=2.$ The fibre is never empty or a singleton, the derivative never degenerates, and quotient seams introduce no boundary endpoints because the calculation uses local lifts. Both inverse branches were displayed explicitly, so no choice principle is used.

Under [F1] the displayed map is $G([t])=\left[\frac{\sin (2\pi t)}{2\pi }\right]\mathrm{.}$ It is well defined because replacing $t$ by an integer translate does not change the sine by [F2]. In local increasing angular coordinates its lifts differ only by integer constants and have derivative $\cos (2\pi t)$ by [F3]; repeated differentiation cycles through sine and cosine, so $G$ is smooth. If $K$ is compact in the target, [F4] makes $K$ closed, hence $G^{-1}(K)$ closed in the compact source and therefore compact. Thus $G$, equivalently $F$, is proper.

The fibre of $[0]$, corresponding to $1\in S^1$, satisfies $\sin (2\pi t)\in 2\pi \mathbb{Z}$. Since $\mathrm{\mid }\sin (2\pi t)\mathrm{\mid }\le 1<2\pi$ by [F2], this is equivalent to $\sin (2\pi t)=0$. The zero-set formula in [F2] gives exactly $[t]=[0]$ or $[t]=[1\mathrm{/}2]$. By [F3] their derivatives are respectively $+1$ and $-1$, so $[0]$ is regular and [F5] gives $\mathrm{deg}(F)=(+1)+(-1)=0,$ although this fibre has two points.

This witnesses the failed unsigned-count conclusion. For comparison, the target value $[1\mathrm{/}2]$ has empty fibre because every lifted value of $G$ has absolute value at most $1\mathrm{/}(2\pi )<1\mathrm{/}2$, and the empty regular-fibre sum again gives zero. The extreme target $[1\mathrm{/}(2\pi )]$ has the singleton preimage $[1\mathrm{/}4]$, but its derivative is $\cos (\pi \mathrm{/}2)=0$, so it is critical rather than a counterexample to the regular-value formula. Quotient seams are handled by local lifts, and every fibre point used above is explicitly listed; no choice principle is used.

The displayed polynomial formula is smooth. At both parameter endpoints, $F_0(x)=H(x,0)=x=H(x,1)=F_1(x)\mathrm{.}$ Thus both endpoint maps are the identity, and each is proper because its inverse image of any compact set is that same compact set.

The compact singleton $\{0\}$ has inverse image $H^{-1}(\{0\})=(\{0\}\times [0,1])\cup (\mathbb{R}\times \{1\mathrm{/}2\})\mathrm{.}$ This inverse image is not compact: for $n\ge 1$, let $U_n=H^{-1}(\{0\})\cap ((-n,n)\times (1\mathrm{/}4,3\mathrm{/}4))$, and let $V=H^{-1}(\{0\})\cap (\mathbb{R}\times ([0,1]\setminus \{1\mathrm{/}2\}))$. These sets are open in the inverse-image subspace and $\{V,U_1,U_2,\dots \}$ covers it, but any finite subfamily misses $(x,1\mathrm{/}2)$ once $\mathrm{\mid }x\mathrm{\mid }$ exceeds every selected index. Hence $H$ is not proper.

The two proper endpoint maps are therefore joined by a nonproper combined homotopy, so the endpoint-only inference fails and [F1]'s hypothesis is indispensable. At $t=1\mathrm{/}2$ the slice is the constant zero map, which pinpoints the degeneracy; at $t=0,1$ it is the identity. The source and compact test set are nonempty, all endpoints and the zero fibre are explicit, and the countable cover is specified by a formula rather than selected, so no choice principle is used.

Suppose $f$ omits a point $y\in N$. Represent $[M]$ by a finite singular cycle $c$. Every simplex of $f_{\mathrm{\#}}c$ lies in $N\setminus \{y\}$, so [F2] gives $q_y(f_{\ast }[M])=0$. But [F1] gives $f_{\ast }[M]=\mathrm{deg}(f)[N]$, hence $0=\mathrm{deg}(f)q_y([N])$. Since $q_y([N])$ is a generator of an infinite cyclic group by [F1], this forces $\mathrm{deg}(f)=0$, contrary to the hypothesis. Therefore $f$ is surjective. This uses the integral homological degree throughout and does not require a comparison with compact-support degree.

For the concrete map, [F3] gives $\mathrm{deg}(P_3)=3\mathrm{\ne }0$, so step 1.1 makes it surjective. The fibres can also be seen directly: for any $[a]\in \mathbb{R}\mathrm{/}\mathbb{Z}$, the three classes $[a\mathrm{/}3]$, $[(a+1)\mathrm{/}3]$, and $[(a+2)\mathrm{/}3]$ map to $[a]$; reducing $3t-a\in \mathbb{Z}$ modulo $3$ proves that these are all the preimages. Hence this example calculates rather than merely naming the surjectivity conclusion.

The nonzero hypothesis is essential: [F3] gives the constant map $P_0$ degree zero, and it omits every point other than $[0]$. In dimension zero, nonempty connected source and target are singletons and the unique map is already surjective, as [F1] records; the relative-chain contradiction in step 1.1 also applies there. Empty or disconnected manifolds lie outside the scalar degree hypotheses, and there are no manifold-boundary endpoints because closed manifolds are boundaryless. Only one omitted point and one representative cycle are used, so no choice axiom is used here.