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Glue two copies of a compact manifold along their boundaries to form its double. A collar gives the double a smooth structure when $W$ is smooth. Folding the copies onto $W$ gives a retraction; in particular, inclusion of either copy is injective on homology. For an oriented manifold, use the opposite orientation on the second copy.

In the Oxford course's convention, a homotopy retract is a retraction $r:X\to A$ such that the inclusion map $i:A\hookrightarrow X$ also satisfies $ir\simeq {\mathrm{id}}_X$. It gives a homotopy equivalence between $A$ and $X$. Unlike the deformation retraction convention used here, the homotopy need not fix $A$ at intermediate times. ^[ref]

Let $X$ be the geometric realization of a finite simplicial complex, or a retract of such a space, and let $f:X\to X$ be continuous. If its Lefschetz number $L(f)$ is nonzero, then $f$ has a fixed point. This includes self-maps of compact manifolds.^[ref]

For a map without fixed points on a finite polyhedron, compactness gives a positive separation between $x$ and $f(x)$. After sufficiently fine simplicial subdivision, a simplicial approximation sends no simplex back onto itself. Its diagonal chain map coefficients vanish, so its alternating trace, and hence $L(f)$, is zero. A retraction reduces the more general stated case to this one.

For example, every continuous self-map of an even-dimensional sphere with mapping degree different from $-1$ has a fixed point, since $L(f)=1+\mathrm{deg}f$. The converse fails: $L(f)=0$ permits fixed points, as the identity map of a circle shows.

Use the mapping cylinder convention $(x,1)\sim f(x)$ for $f:X\to Y$. Its bottom copy of $X$ is a retract precisely when $f$ has a left homotopy inverse $g:Y\to X$. A retraction restricted to $Y$ supplies $g$, and its values along the cylinder supply the homotopy from ${\mathrm{id}}_X$ to $gf$. Conversely use that homotopy on the cylinder and $g$ on $Y$; agreement at the attached end makes the map descend by the universal property of the quotient topology.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

A retraction from a topological space $X$ onto a subset $Y$ is a continuous map $r:X\to Y$ satisfying $r(y)=y$ for every $y\in Y$. Equivalently, $r\circ \iota ={\mathrm{id}}_Y$, where $\iota :Y\hookrightarrow X$ is the inclusion.

The no-retraction theorem for the finite-dimensional closed unit ball saysIndeed, if $r$ were such a retraction, $F(x)=-r(x)$ would be a continuous map from ${\mathbb{B}}^n$ into itself. The Brouwer fixed-point theorem would give $x=F(x)=-r(x)$. Since $|r(x)|=1$, $x$ lies on the boundary, where $r(x)=x$. Thus $x=-x$, contradicting $|x|=1$. Here the Brouwer fixed-point theorem asserts that every continuous map of a finite-dimensional closed unit ball into itself has a fixed point.

The retraction of a Hilbert ball onto its sphere exists. In the real Hilbert space ${\ell }^2$, let $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$ be the unilateral shift operator, and let $e_0=(1,0,\dots )$. For $x$ in the closed unit ball $\mathbb{B}$, defineThe orthogonality of $e_0$ and $Sx$ gives $\Vert F(x){\Vert }^2=1$. Thus $F$ is a continuous map into the unit sphere. It has no fixed point: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\ge 0$. A constant sequence in ${\ell }^2$ must be zero, whereas $F(0)=e_0\ne 0$.

Put $d=x-F(x)$, which is nonzero everywhere, and defineThis is a continuous map on the whole closed unit ball, including zero. To check its range, observe that $⟨F(x),d⟩=⟨F(x),x⟩-1$, soIf $\Vert x\Vert =1$, then $\Vert x-F(x){\Vert }^2=2(1-⟨F(x),x⟩)$, hence $\alpha (x)=1$ and $r(x)=x$. Therefore $r:\mathbb{B}\to \partial \mathbb{B}$ is a retraction.

Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the unit sphere. In factThe construction illustrates why the finite-dimensional hypothesis in the no-retraction theorem matters. The closed unit ball of ${\ell }^2$ is not a compact set in the norm topology.

The triangular-grid form is Sperner's lemma: subdivide a triangle into small triangles meeting edge to edge, label its vertices $1,2,3$, and forbid label $i$ on the side opposite corner $i$. The corner labels are thereby forced. An odd number of small triangles, in particular at least one, carry all three labels.

Here is the parity proof. Count incidences of a small triangle with an edge whose endpoint labels are $1,2$. On the outer boundary these edges occur only on the side joining corners $1,2$; traversing that side begins with label one and ends with label two, so the number of changes of label is odd. Interior edges contribute twice to the incidence count. A small triangle with all three labels contributes once; a triangle having only labels one and two contributes twice if it has both, and all other triangles contribute zero. Reduction modulo two therefore proves the assertion.

To obtain the continuous Brouwer fixed-point theorem on the triangle, use barycentric coordinates and a continuous map $F$ from the triangle to itself. If a grid vertex is not already fixed, label it by an index $i$ with $x_i>F_i(x)$, which exists since both coordinate sums are one. A point on the side $x_i=0$ cannot receive label $i$, so the discrete lemma applies. On successively finer grids take a fully labelled small triangle. Compactness gives a subsequence converging to one point $x$; all three vertices converge to it because the mesh tends to zero. Passing to the limit in the three label inequalities gives $x_i\ge F_i(x)$ for every $i$. Equal coordinate sums force equality throughout, hence $F(x)=x$.

A closed disk is homeomorphic to a triangle: radial coordinates from an interior point, scaled by the distance to the boundary in each direction, give such a homeomorphism. Conjugation therefore gives the same fixed-point assertion on the disk. If $r$ were a continuous retraction from the unit disk to its boundary, define $F(x)=-r(x)$. A fixed point would have $|x|=1$, hence $r(x)=x$ and $x=-x$, impossible on the unit circle. Thus no continuous disk-to-boundary retraction exists. The continuous conclusion has been deduced from the grid parity argument, not used to prove it.

The Brouwer fixed-point theorem says that every continuous map from a closed Euclidean ball to itself has a fixed point. For the closed disc $D\subset {\mathbb{R}}^2$ its equivalent no-retraction theorem says that there is no continuous $r:D\to \partial D$ satisfying $r(z)=z$ on $\partial D$.

Assume the fixed-point theorem and suppose such a retraction exists. Then $F(x)=-r(x)$ maps $D$ continuously into itself. A fixed point must lie on the boundary, where $r(x)=x$, giving $x=-x$, impossible on the unit circle.

Conversely suppose $F:D\to D$ is continuous and has no fixed point. From $F(x)$ draw the ray through $x$ and let $r(x)$ be its exit point on the circle. To check continuity rather than merely relying on the picture, put $v=x-F(x)\ne 0$. The exit point is $F(x)+t(x)v$, whereThis is continuous, lies on the circle and has $t(x)\ge 1$. On the boundary the exit point is $x$, so $r$ is a retraction, contradicting the assumed theorem. Hence the two disc formulations are equivalent; the same argument works for a ball in any dimension.

The planar Brouwer fixed-point theorem says that every continuous map from the closed disk $D$ to itself has a fixed point. If a retraction $r:D\to \partial D$ existed, $x\mapsto -r(x)$ would be a continuous self-map of $D$. At a fixed point $x=-r(x)$, $x$ lies on the boundary, where $r(x)=x$; this would give $x=-x$, impossible on the unit circle.

Conversely, suppose a continuous self-map $f:D\to D$ has no fixed point. For each $x$, take the ray starting at $f(x)$ through $x$, and let $r(x)$ be its exit point on $\partial D$. The denominator $|x-f(x)|$ never vanishes, and solving the quadratic intersection with the circle shows that the exit point depends continuously on $x$. More explicitly, with $v=x-f(x)$,The ray exits beyond or at $x$, so $\lambda \ge 1$, and when $x\in \partial D$ its exit point is $x$. Hence $r$ is a retraction. Nonexistence of a boundary retraction and the planar fixed-point theorem are equivalent.

Suppose first that a retraction $R:D\to C$ exists. The continuous map $p\mapsto -R(p)$ takes $D$ to itself, so the Brouwer fixed-point theorem gives $p=-R(p)$. Such $p$ lies on $C$, where $R(p)=p$, contradicting $p=-p$.

Conversely, suppose a continuous map $T:D\to D$ has no fixed point. For each $p$, follow the ray from $T(p)$ through $p$ to its last intersection with $C$. This defines a retraction. More explicitly, put $a=p-T(p)\ne 0$ and setThe positive-root formula is continuous, $t(p)\ge 1$, and $R(p)$ is on the unit circle. If $p\in C$, the last intersection is $p$ itself, so $R(p)=p$. Thus the nonexistence of a retraction rules out a fixed-point-free $T$, proving the equivalence.

No such map exists. If $H$ fixed the boundary, it would satisfy the hypothesis of part (iii) for every $\epsilon >0$. Choose $0<\epsilon <1-|q|$. Then part (iii) would give $q\in H(B)$, contradicting the stipulated omission of $q$. Equivalently, radial projection from the omitted point would turn $H$ into a forbidden retraction.

All homology and cohomology in this solution use ${\mathbb{F}}_2$ coefficients, so no orientation hypothesis is necessary. The Poincare-Lefschetz duality degrees are complementary:The starred notation in the question must be understood with this degree reversal.

Turn the handle decomposition upside down. An absolute $k$-handle becomes a relative $(n-k)$-handle based on $\partial W$. The absolute cellular boundary coefficient counts intersections of an attaching sphere with a belt sphere. In the reversed handle decomposition, the same intersections are counted in the opposite order. Over ${\mathbb{F}}_2$ the two incidence matrices are transposes, with no sign ambiguity. Thus the absolute chain complex is the complementary-degree dual of the relative chain complex. Taking homology and using the universal coefficient theorem for cohomology over a field proves the displayed isomorphism. This is mod-two handle duality.

To obtain a perfect pairing from the assumed closed case, form the double of a manifold $D(W)$ from two copies of $W$. The fold map $D(W)\to W$ is a retraction, so the inclusion of either copy induces an injection on homology. If $a\ne 0$ in $H_k(W)$, its image in $H_k(D(W))$ is nonzero. Closed-manifold mod-two Poincare duality supplies a class $z\in H_{n-k}(D(W))$ with intersection $1$ against that image. Cut a representative of $z$ along the boundary and retain its part in the first copy; it defines $b\in H_{n-k}(W,\partial W)$. Equivalently, apply the quotient map $D(W)\to D(W)/W_{-}\cong W/\partial W$ and excision. Intersections with a representative of $a$ in the interior are unchanged, so $a\cdot b=1$. This proves nondegeneracy in the first variable. Mod-two handle duality gives equal finite dimensions for the two spaces, hence nondegeneracy in the second variable as well.

For the three-dimensional conclusion, setThe long exact sequence in relative homology says $L=\mathrm{im}\partial$, where $\partial :H_2(W,\partial W)\to V$. The closed-surface intersection pairing on $V$ is nondegenerate. A collar calculation gives the adjoint identityIndeed, push $x$ slightly into the collar: intersections with the relative surface representing $b$ correspond to its boundary intersections with $x$. By the absolute-relative perfect pairing already proved, $x$ is orthogonal to every $\partial b$ precisely when $i_{\ast }x=0$. Therefore $L^{\perp }=L$.

For any finite-dimensional space with a perfect pairing, $\dim L+\dim L^{\perp }=\dim V$. Hence the half-lives-half-dies theorem givesBut $H_1({\mathbb{RP}}^2;{\mathbb{F}}_2)\cong {\mathbb{F}}_2$ has dimension one. It cannot be the entire boundary of a compact three-manifold, because the displayed dimension would be $1/2$. This excludes nonorientable three-manifolds as well as orientable ones.

The topological degree measures an oriented net number of sheets of a map. Its simplest integer-valued setting is a map $f:M\to N$ between connected oriented closed manifolds of the same positive dimension $n$. The hypotheses matter: an integer sign count needs orientations; compactness prevents preimages from escaping; and the same dimension makes a regular fiber discrete. A fundamental class gives the definition for continuous maps:Here $H_n(M;\mathbb{Z})$ and $H_n(N;\mathbb{Z})$ are generated by their chosen fundamental classes. Reversing either orientation changes the sign of the degree of a map between oriented manifolds. Disconnected sources contribute a sum over components; disconnected targets require a degree for each target component rather than one universal integer.

For a smooth map, choose a regular value $y$. The Sard theorem guarantees plentiful choices. Each point of $f^{-1}(y)$ has an invertible derivative, so the inverse function theorem makes it an isolated point. The fiber is finite by compactness of $M$. An oriented coordinate chart defines $\mathrm{sgn}\det d{f}_x$, and the degree as a sum of local degrees becomesAn empty fiber contributes zero. The sign is independent of the chosen oriented charts. Thus the mapping degree counts sheets with signs, not simply the number of preimages: orientation-preserving and orientation-reversing sheets cancel.

A differential-form argument both proves that this count is independent of $y$ and relates it to the topological definition. Around a regular value, take a small neighborhood $V$ whose full preimage is a disjoint union of neighborhoods mapped diffeomorphically onto $V$. Compactness rules out additional preimages approaching $V$ from elsewhere. Choose a smooth top-degree form $\eta$ supported in $V$, with ${\int }_N\eta =1$. The change of variables formula givesAlthough $\eta$ need not be everywhere positive, it represents the same normalized top cohomology class as any normalized volume form. The top de Rham cohomology of a compact connected oriented manifold says that two top forms with equal integrals differ by $d\beta$. Their pullback integrals agree, because the Generalized Stokes theorem gives ${\int }_{M}d(f^{\ast }\beta )=0$. Therefore the degree by integration of a pullback volume form isfor every smooth top form $\eta$. Pairing de Rham cohomology with fundamental classes identifies this integer with the homological definition. This proof also shows that every regular value gives the same signed sum.

The homotopy invariance of mapping degree makes it stable under continuous deformation. Homologically this follows from homotopic maps inducing the same map on homology. For a smooth homotopy $F:M\times [0,1]\to N$, a direct proof uses $d\eta =0$ and the Generalized Stokes theorem on the cylinder to getwith the cylinder orientation chosen to give the indicated boundary signs. Smooth approximation extends the geometric description to continuous maps, while the fundamental class definition already applies without differentiability. The multiplicativity of mapping degree follows from functoriality and givesA constant map has degree zero when $n>0$. If the mapping degree is nonzero, the map is onto: a point outside its image is a regular value with empty preimage. An orientation-preserving diffeomorphism has degree $1$, and an orientation-reversing one has degree $-1$. An orientation-preserving $k$-sheeted covering map has degree $k$; the signs of sheets must be included for other orientation conventions.

Examples show both the strength and the limits of the invariant. On the circle, $e^{i\theta }\mapsto e^{ik\theta }$ has degree $k$, including negative integers and zero; it is the winding number. On the sphere $S^n$, an ambient reflection has degree $-1$, and the antipodal map has degree $(-1{)}^{n+1}$, because the determinant of $-I$ on the ambient ${\mathbb{R}}^{n+1}$ is $(-1{)}^{n+1}$. On the torus, an integer matrix $A$ induces a map of degree $\det A$, obtained by pulling back the constant top form. However, degree does not classify general manifold maps: the identity and an integer shear on the two-torus both have degree one but induce different maps on $H_1$, so are not homotopic.

The mapping degree gives useful obstruction and existence arguments. There is no retraction of a closed ball onto its boundary sphere: such a retraction would make the degree-one identity of the boundary null-homotopic, since it would extend across a contractible ball, contradicting homotopy invariance and the degree-zero constant map. This is the degree obstruction underlying the Brouwer fixed-point theorem. Also an everywhere nonzero tangent vector field on an even-dimensional sphere would, after normalization, produce the homotopy $\cos (\pi t)x+\sin (\pi t)v(x)$ from the identity to the antipodal map. Orthogonality makes every value a unit vector, but the endpoint degrees are $1$ and $-1$, impossible. This proves the corresponding Hairy ball theorem obstruction.

There are appropriate extensions rather than an unrestricted integer degree for every pair of manifolds. For a proper map between connected oriented manifolds without boundary of equal dimension, the regular-value count remains finite and is invariant; compactly supported top forms or locally finite homology replace the compact fundamental-class description. Without orientations, degree modulo two counts preimages modulo two and uses mod-two fundamental classes. For manifolds with boundary one uses relative mapping degree for maps of pairs, keeping the boundary away from the chosen target value. Maps of unequal dimensions, or nonproper maps where preimages can escape to infinity, do not in general admit this same signed sheet-count invariant. The central principle is an integer signed count stable under homotopy, with its hypotheses explicitly preserved.

Sperner's lemma says that a triangulated triangle whose vertices are labelled $1,2,3$, with each outer vertex having its own label and each boundary edge using only its endpoint labels, contains a small triangle with all three labels. In fact the number of such triangles is odd. Count incidences of small triangles with edges labelled $1,2$. Interior edges contribute twice. On the outer $1,2$ edge, the number of changes between these labels is odd; the other two outer edges contribute zero. A small triangle contributes one incidence if it has all three labels, two if it uses both $1,2$ but not $3$, and zero otherwise. The total parity proves the assertion.

The resulting no-retraction theorem is: there is no continuous retraction of a closed disc onto its boundary. Work first with a closed triangle $P$. If $r:P\to \partial P$ fixes the boundary, colour each vertex of increasingly fine triangulations by an index of a largest barycentric coordinate of $r(v)$. The largest coordinate is at least $1/3$; on the boundary the colouring satisfies Sperner's lemma. Choose a tricoloured small triangle in each triangulation. By compactness and shrinking mesh, a subsequence of their three vertices converges to one point $x$. Continuity forces all three barycentric coordinates of $r(x)$ to be at least $1/3$, so $r(x)$ is the triangle's centre, contradicting $r(x)\in \partial P$. A homeomorphism of a disc with $P$ proves the disc assertion.

Brouwer fixed-point theorem states that every continuous self-map $F$ of a closed disc has a fixed point. If it had none, the ray from $F(x)$ through $x$ would meet the boundary beyond $x$, continuously in $x$. That intersection defines a retraction, since a boundary point is already the outgoing intersection. This contradicts the no-retraction theorem.

Finally fix $y\in {\mathbb{R}}^2$ and define $F_y(x)=y+x-g(x)$. The displacement bound implies $\Vert F_y(x)-y\Vert \le K$, so $F_y$ maps the closed radius-$K$ disc centred at $y$ into itself. Brouwer fixed-point theorem gives $F_y(x)=x$, which is exactly $g(x)=y$. Since $y$ was arbitrary, $g$ is surjective.

Put $A={\Sigma }_h^{\partial }$. This surface with boundary deformation retracts onto a wedge of $2h$ circles, so $H^1(A;\mathbb{R})$ has dimension $2h$ and $H^2(A;\mathbb{R})=0$. If $r$ were a retraction, ${\iota }^{\ast }{r}^{\ast }=\mathrm{id}$ would make $r^{\ast }$ injective. Its image $W\subset H^1({\Sigma }_g;\mathbb{R})$ would have dimension $2h$.

Every pair of classes $u,v$ in $H^1(A;\mathbb{R})$ has $u⌣v=0$, since $H^2(A;\mathbb{R})=0$. Naturality givesThus $W$ is an isotropic subspace of a symplectic vector space for the nondegenerate skew Poincare duality pairing on the $2g$-dimensional space $H^1({\Sigma }_g;\mathbb{R})$. The stated linear-algebra bound gives $2h\le g$. Hence

The bound is sharp. Double $A$ along its boundary: $D(A)=A{\cup }_{\partial A}A$ is the closed oriented surface of genus $2h$. Identify each copy with $A$ and fold them onto one copy. The two maps agree on the joining circle, so they give a continuous retraction fixing the first copy pointwise. This includes $h=0$, where the double of a disc is a sphere.

More generally, for $g\ge 2h$, add $g-2h$ handles in the interior of the second copy. Pinch those extra handles onto their connecting point, keeping the boundary fixed, and then fold onto $A$. This is still a retraction. In fact the construction works for any embedding in the question: the connected complement has one boundary component and genus $g-h$ by Euler-characteristic additivity, and the classification theorem for surfaces identifies it, relative to that boundary, with a copy of $A$ having $g-2h$ additional handles. Thus the exact existence criterion is $2h\le g$, as expressed by retraction onto a punctured oriented surface.

No, it cannot deformation retract onto a graph. Each sphere contributes a degree-$2$ homology class independent of the other two. Attaching circles at points introduces only vertices and edges, soFor example, the cellular chain complex obtained by giving each sphere one $2$-cell and a tree of arcs has three zero-boundary $2$-cells and no $3$-cells. Every graph has zero degree-$2$ homology. A deformation retraction would induce a homotopy equivalence and an isomorphism on homology, which is impossible. An ordinary retraction has no such requirement.

Yes, it retracts onto a graph. On each of the three spheres, choose an arc joining the two marked points. A sphere retracts onto this arc: identify it with the vertical meridian parameterized by the height coordinate, and send every point of a given height to the meridian point at that height. This map is continuous and fixes the meridian. Apply these three retractions and fix the lifted circles pointwise; the maps agree at all attachment points. The resulting connected graph has six vertices and nine edges, so its fundamental group is free of rank $\boxed{9-6+1=4}$.

Label the edges so that barycentric coordinates on $T$ satisfySuppose for contradiction that $A\cap B\cap C=\varnothing$. The distance from a point to a closed set gives continuous nonnegative functions $d_A,d_B,d_C$. Their sum never vanishes, while at every point at least one of them vanishes because $A\cup B\cup C=T$. Henceis a continuous map from $T$ to its boundary in barycentric-coordinate space. Moreover, $F(I)\subseteq I$, $F(J)\subseteq J$, and $F(K)\subseteq K$. On each edge, the straight-line homotopy between $F$ and the identity remains in that edge, so $F{|}_{\partial T}$ has winding number one. On the other hand, a map from the whole triangle to its boundary makes its boundary restriction null-homotopic, and hence gives winding number zero. This contradiction proves the three-set covering lemma:

If a retraction $f:D\to \partial D$ fixed every boundary point, choose a homeomorphism from $T$ to $D$ taking its three edges to three consecutive closed arcs of $\partial D$ with empty triple intersection. The inverse images under $f$ of those arcs would be closed, would cover $D$, and would contain the corresponding boundary arcs. Transporting them to $T$ would contradict the result just proved. Therefore

For the final statement, use compactness of the three closed arcs inside the corresponding open sets. A finite open cover of a compact metric space admits a closed shrinking, and the shrinking can be chosen to preserve specified compact subsets already lying in the respective open sets. Thus there are closed setswhich cover $D$ and contain $\alpha ,\beta ,\gamma$, respectively. The disc version of the three-set covering lemma, obtained from the triangle by a homeomorphism taking its edges to the three arcs, gives a point in $A\cap B\cap C$. Since these sets lie in the original open sets,

Let $i:A\hookrightarrow X$ be the inclusion map and $r:X\to A$ the retraction. Since $X$ is contractible, choose a contraction of a topological spaceThe function compositionis a continuous map. Its endpoints satisfyConsequently $K$ is a homotopy from the identity map of $A$ to a constant function. This provesThe contracting point of $X$ need not belong to $A$: applying the retraction sends it to the contracting point $r(x_0)$ of $A$. This is the retract of a contractible space principle.