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Let $\{O_i\}$ be an open cover of $Z$ in its subspace topology. Write $O_i=Z\cap U_i$ with each $U_i$ open in $X$. Since $Z$ is a closed set, $X\setminus Z$ is open, and $\{U_i\}\cup \{X\setminus Z\}$ is an open cover of the compact space $X$. A finite subcover restricts to a finite subcover of $Z$. Hence the closed subset $Z$ is compact.

If $\{V_j\}$ is an open cover of $f(Z)$ in its subspace topology, the inverse images under the restricted continuous map $f{|}_Z$ form an open cover of $Z$. Take a finite subcover; the corresponding $V_j$ cover $f(Z)$. Thus the continuous image $f(Z)$ is compact. Neither argument assumes a Hausdorff space or asserts that a compact set must be closed set in an arbitrary topological space.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

The dynamical proof of Hindman's theorem gives the following result. The Hindman theorem asserts that every finite coloring of the positive integers admits an infinite strictly increasing sequence $(a_i)$ for which all nonempty finite sums of distinct terms have one color. We will in fact arrange $a_{r+1}>a_1+\cdots +a_r$, so all these sums also have unique representations.

Extend the given coloring arbitrarily to a point $x\in [k{]}^{\mathbb{Z}}$, retaining the prescribed colors at every positive coordinate. Let $X$ be its forward orbit closure under the left shift. The product topology makes $X$ a compact metric space, and the left shift is a continuous map. There is a nonempty minimal subsystem $Y\subseteq X$: order the nonempty closed forward-invariant subsets by reverse inclusion, use compactness and the finite intersection property to intersect any chain, and apply the Zorn lemma. Minimality also implies $\mathcal{L}(Y)=Y$. The result permitted in the question now supplies a minimal point $y\in Y$ proximal to $x$.

We need the joint return lemma for a proximal minimal pair, which we prove here. It suffices to consider an open neighborhood $U$ of $y$ in $X$. Choose an open neighborhood $V$ of $y$ with $\overline{V}\subseteq U$. Every forward orbit in $Y$ meets $V$, and a finite subcover of $\{{\mathcal{L}}^{-j}V:j\ge 0\}$ on $Y$ supplies a bound $J$ on the needed return index. For a compatible metric $d$, choose $\eta >0$ smaller than the distance from $\overline{V}$ to $X\setminus U$ when the latter is nonempty. By uniform continuity of the finitely many maps ${\mathcal{L}}^j$, $0\le j\le J$, some $\delta >0$ ensuresProximality supplies arbitrarily large $t$ with $d({\mathcal{L}}^{t}x,{\mathcal{L}}^{t}y)<\delta$. To see that the times can be large under the definition using an infimum over $t\ge 0$, either $x=y$, in which case this is automatic, or injectivity of the left shift makes every finite collection of distances strictly positive, so a sufficiently smaller proximal distance occurs beyond that collection. Some $j\le J$ has ${\mathcal{L}}^{t+j}y\in V$. Then ${\mathcal{L}}^{t+j}x\in U$ also. Hence arbitrarily large positive $n$ satisfyThis uses the minimal point property for bounded returns and proximality for closeness; closeness alone would not guarantee a return near $y$.

Put $q=y(0)$. Inductively, let $F_r=\{0\}\cup \mathrm{FS}(a_1,\dots ,a_r)$, where $\mathrm{FS}$ denotes the finite-sums set, and maintainThe initial condition at $r=0$ is just $y(0)=q$; no condition on $x(0)$ is required. The cylinder set $U_r=\{z:z(s)=q\text{ for every }s\in F_r\}$ is a neighborhood of $y$. Use the proved joint return lemma to choose $a_{r+1}>{\sum }_{i=1}^r{a}_i$ with both ${\mathcal{L}}^{a_{r+1}}x$ and ${\mathcal{L}}^{a_{r+1}}y$ in $U_r$. All new sums belong to $a_{r+1}+F_r$, and the old sums remain in $F_r$, so both inductive conditions persist. Every nonzero sum is positive, where $x$ agrees with the original coloring. ThereforeThis proves the Hindman theorem using only the permitted proximal-minimal existence result and the compactness and return arguments supplied above.

The relevant Penrose singularity theorem states: a connected time-oriented four-dimensional globally hyperbolic spacetime with a noncompact Cauchy hypersurface, a nonempty compact orientable boundaryless trapped surface, and the null convergence condition is future null-geodesically incomplete. Under the Einstein field equations, the null energy condition supplies the curvature hypothesis. A trapped surface here is spacelike and has both future normal null expansions strictly negative. The conclusion is an incomplete null geodesic, not necessarily a divergent curvature invariant at an identifiable point.

Suppose, for contradiction, that every future null geodesic is complete. Write $S$ for the trapped surface andThe equality follows because the causal future of a compact set is closed in a globally hyperbolic spacetime. This future horismos is a closed achronal boundary. It is nonempty: the restriction of a Cauchy time function to compact $S$ has a minimum, and a point at that minimum cannot be chronologically preceded by another point of $S$.

Normalize the two future null normal directions along $S$ against a smooth future timelike field, fixing the affine scale continuously. Their initial null expansions are continuous and strictly negative. Compactness of the normalized normal bundle gives a uniform $c>0$ with ${\theta }_{\pm }\le -c$ everywhere on $S$. The preceding null focusing theorem forces a conjugate point to a spacelike surface along each normal generator within affine length $L=2/c$.

By the supplied boundary-generator result, every point of $E^{+}(S)$ is reached by an orthogonal future null geodesic which has no earlier conjugate point. No such boundary generator can remain on the boundary beyond its first focal point. Every point of $E^{+}(S)$ therefore lies inThe parameter set is compact. Future null completeness makes its geodesic flow defined throughout this common finite interval, and smooth dependence on initial data makes its image $K$ compact. Since $E^{+}(S)$ is closed and contained in $K$, it is compact. This is the compactness of the future horismos of a trapped surface step; a merely pointwise finite bound would not suffice without compactness and uniform normalization.

Choose a smooth complete timelike vector field, obtained if necessary by positive rescaling against a complete auxiliary Riemannian metric. Its inextendible integral curves of a vector field meet a chosen smooth Cauchy hypersurface $C$ exactly once. Projection along those curves defines a continuous map ${\pi }_C:E^{+}(S)\to C$. The achronal set property makes this map an injective function: two boundary points on the same timelike integral curve of a vector field would be timelike related.

An achronal boundary is a topological hypersurface without boundary, even at nonsmooth generator junctions; the submanifold property allowed in the question supplies this fact. Thus domain and codomain are both three-dimensional topological manifolds without boundary. Invariance of domain makes ${\pi }_C(E^{+}(S))$ open in $C$. Compactness makes it closed in the Hausdorff space $C$, and it is nonempty. Connectedness of the spacetime gives connectedness of $C$, so this image is all of $C$. That would make $C$ compact, contradicting the noncompact Cauchy hypersurface hypothesis. The completeness assumption is false, proving the theorem.

The noncompactness hypothesis and strict trapping are essential to this version. Global hyperbolicity cannot simply be omitted in the compactness and projection steps, and nonpositive initial expansion with zeros does not supply the uniform focusing bound used here.

For arbitrary $y\in {\mathbb{R}}^2$, the continuous map $g_y(x)=g(x)-y$ satisfies $\Vert g_y(x)-x\Vert \le K+\Vert y\Vert$. Applying the preceding Brouwer fixed-point theorem argument with this positive bound gives $g_y(t)=0$. Hence $g(t)=y$, and $g$ is surjective.

Let $u(t)=\gamma (t)/|\gamma (t)|$, a continuous map into the unit circle. By uniform continuity, partition $[0,1]$ so finely that on each interval beginning at $t_j$, $|u(t)/u(t_j)-1|<1$. This ratio lies in the right half-plane and has a continuous complex argument with value zero at $t_j$. Starting from ${\theta }_0$, add these local arguments successively, matching endpoint values. This constructs a continuous real lift with $u(t)=e^{i\theta (t)}$.

Any two such lifts with the same initial value differ by a continuous function taking values in $2\pi \mathbb{Z}$, and hence coincide. Since the curve closes, $e^{i(\theta (1)-\theta (0))}=1$. Define its winding number byChanging the chosen initial argument adds a constant multiple of $2\pi$ to the whole lift, so the winding number is independent of that choice. No differentiability of the curve is required.

Suppose such an $f$ exists. Orthogonality and unit lengths makea continuous map into the unit sphere, since its squared norm is one. This is a homotopy from the identity to the antipodal map. Their degrees would coincide, but for even $n\ge 2$ they are respectively $1$ and $(-1{)}^{n+1}=-1$, a contradiction. Thus no such continuous map exists for even $n$. At $n=0$ there is not even a unit vector orthogonal to either point of $S^0\subset \mathbb{R}$, so nonexistence holds there as well. This is a sphere form of the Hairy ball theorem.

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

A topological space $X$ is path-connected if, for every $a,b\in X$, there is a continuous mapThis continuous path must stay within $X$ for its entire duration. Its parameter need not represent physical time, and continuity does not require a derivative or finite arc length. ^[ref]

In the closed interval $[0,1]$, any $a,b$ are joined by $\gamma (t)=(1-t)a+tb$. For $a=1/4$ and $b=3/4$, this becomes $\gamma (t)=1/4+t/2$, so $\gamma (0)=1/4$, $\gamma (1/2)=1/2$, and $\gamma (1)=3/4$. All its values stay in $[0,1]$. In the union $D=[-2,-1]\cup [1,2]$, however, a continuous path from $-1$ to $1$ would attain $0$ by the intermediate value theorem, although $0\notin D$. Thus $D$ is not path-connected.

Every path-connected space is a connected space. Proof outline: if $X=U\cup V$ admitted a separation of a topological space, choose $a\in U$ and $b\in V$. A continuous path between them would have nonempty open inverse images ${\gamma }^{-1}(U),{\gamma }^{-1}(V)$ with empty intersection and union $[0,1]$. This contradicts the connectedness of that closed interval.

The converse fails: the topologist's sine curve is connected but has two path components. Path connectedness is useful when we need an actual continuous path of allowed positions, rather than only the absence of a separation of a topological space. For instance, when modelling allowed positions as a topological space, membership in the same path component answers whether one can move continuously between two positions while remaining in that topological space.

A continuous map on a nonempty topological space is point-transitive if it has a dense forward orbit. This is one convention for topological transitivity, but on spaces with isolated points it must be distinguished from the open-set intersection definition. On a nonempty compact metric space, the positive-time open-set intersection property implies point transitivity: for each member $U_j$ of a countable basis of a topology, ${\bigcup }_{n\ge 1}{f}^{-n}(U_j)$ is open and dense, and the Baire category theorem supplies a point in their intersection. Conversely, a surjective point-transitive continuous map has every tail of its dense orbit dense, since $f^k(X)=X$ and continuity sends its orbit closure into the closure of that tail. The map $f(1/j)=1/(j+1)$, $f(0)=0$ on $\{0\}\cup \{1/j:j\ge 1\}$ is point-transitive but fails the open-set intersection property.

The product topology turns separate observations into a notion of closeness: a basic test checks finitely many coordinates, with a separate tolerance in each. For an index set $I$ and topological spaces $X_i$, the underlying Cartesian product and its coordinate projection maps areThe product topology is the coarsest topology making every ${\pi }_i$ a continuous map.^[ref]

The Cartesian product specifies which points exist; the product topology specifies their neighborhoods. Finite observations motivate its finite-coordinate basis for the product topology, while convergence in a product space and continuity into a product space explain why it is useful. The compactness result for this construction is Tychonoff's theorem.

In a compact metric space with a continuous map $T$, points $x,y$ are proximal ifThis property is independent of the compatible metric. If $T$ is injective and $x\ne y$, arbitrarily small proximal distances must occur at arbitrarily large times, since each finite collection of distances is strictly positive. Proximal points need not be equal or have dense orbits.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.

The quotient topology turns an instruction to glue points together into a precise topological space. Start with a topological space $(X,{\tau }_X)$ and an equivalence relation $\sim$. The quotient set $X/\sim$ consists of equivalence classes, and its canonical projection of a quotient set isThe quotient topology is defined by the rule using preimages of open sets:Thus a subset of the quotient space is open if and only if its entire preimage is open in $X$. Each fiber $q^{-1}(\{[x]\})$ is the whole equivalence class $[x]$, so the preimage includes every point glued to any of its members.^[ref]

This is the finest topology making $q$ a continuous map: every topology with that property can contain only subsets whose preimages are open. It really is a topology, because preimages preserve the empty set, the whole set, arbitrary unions, and finite intersections. The gluing example below shows why checking a complete preimage matters, especially at the seam.

The quotient topology also makes continuous maps out of a glued space easy to construct. A continuous map $f:X\to Z$ descends to a unique continuous map $\bar{f}:X/\sim \to Z$ exactly when it is constant on every fiber of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar{f}([x])=f(x)$; the equivalence class representative cannot affect the answer. For every open set $V\subseteq Z$,The right side is open, so the quotient topology makes ${\bar{f}}^{-1}(V)$ open. Conversely, if $f=\bar{f}\circ q$ for a continuous map $\bar{f}$, then composition of continuous functions makes $f$ continuous, and equal equivalence classes force equal values. This is the universal property of the quotient topology. ^[ref]

A surjective continuous map is not automatically a quotient map. On the same two-point set $D=\{a,b\}$, take the identity map from the discrete topology to the indiscrete topology. It is continuous and surjective, but $\{a\}$ is not open in the target even though its preimage is open in the domain. The quotient topology for this identity map would be the discrete topology. Thus continuity supplies only one direction of the rule using preimages of open sets; a quotient map requires both.

For a continuous map $f:M\to N$ of nonzero mapping degree between closed connected oriented manifolds of the same dimension, $f^{\ast }:H^q(N;\mathbb{Q})\to H^q(M;\mathbb{Q})$ is injective in every degree. Pair a nonzero class $a$ with a complementary class $b$ using Poincare duality. Naturality gives $⟨f^{\ast }a⌣f^{\ast }b,[M]⟩=(\mathrm{deg}f)⟨a⌣b,[N]⟩\ne 0$, so $f^{\ast }a\ne 0$. Over coefficients of positive characteristic the argument requires that the degree remain nonzero in that field.

The continuous map projecting a real number onto the closed interval $[a,b]$, with $a\le b$. It equals $a$ below the interval, equals the input inside it, and equals $b$ above it. It is nondecreasing and has Lipschitz constant one.

A singular $n$-simplex in a space $X$ is a continuous map $\sigma :{\Delta }^n\to X$ from the standard $n$-simplex. Its alternating restrictions to the faces form its boundary.

A continuous map from a smooth manifold to ${\mathbb{R}}^d$ can be approximated by a smooth map with any prescribed positive continuous pointwise error bound. In coordinate charts, smoothing by local convolution and then combining with a smooth partition of unity gives the approximation. On a compact subset of an open chart domain, a smooth cutoff allows the resulting straight-line homotopy to be supported in that domain. This permits the Sard theorem to be used after approximation; the original continuous map can be space-filling and need not omit a point.

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.

If a continuous map $g:{\mathbb{R}}^n\to {\mathbb{R}}^n$ has $\Vert g(x)-x\Vert \le K$, then it is onto. For a target $y$, choose $R\ge K+\Vert y\Vert$, $R>0$, and map the unit ball to itself by $f(x)=x-(g(Rx)-y)/R$. A fixed point gives $g(Rx)=y$.