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Algebraic topology translates questions about topological spaces and continuous maps into questions about groups and other structures from algebra. A topological invariant compresses the enormous amount of data in a space into something one can calculate. For example, the fundamental group records which based loops can be continuously deformed into one another. A filled disk has trivial fundamental group, whereas a circle has fundamental group $\mathbb{Z}$: an integer remembers how many times a loop goes around.

This compression also respects maps. By functoriality of the fundamental group, a continuous map preserving the chosen basepoints induces a group homomorphism; successive maps induce successive group homomorphisms. Thus geometry supplies arithmetic constraints. A proposed homotopy equivalence must induce an isomorphism, so incompatible fundamental groups rule it out before anyone attempts to construct the deformation.

The beauty is that a whole annulus can retain exactly the same homotopy type as its inner circle. Its thickness disappears under a deformation retraction, while the obstruction to shrinking a loop survives. Equal fundamental groups alone do not establish a homeomorphism or a homotopy equivalence between arbitrary topological spaces.

This turns a visual question into a reusable test. Suppose a continuous map sends the inner circle of an annulus once around another circle. Could it extend over a filled disk spanning that boundary? Such an extension would supply a homotopy shrinking the image loop to a point. Its winding number would have to change from one to zero, which homotopy invariance of winding number forbids. Algebra rules out the extension even when no particular formula for it has been proposed.

In the standard free basis of a group $a,b$ for the fundamental group of a once-punctured torus, its oriented boundary loop represents $ab{a}^{-1}{b}^{-1}$ up to orientation and conjugation. The square model proves this: radial deformation retraction takes the inner boundary once around the four sides, whose orientations read $a,b,a^{-1},b^{-1}$. Attaching a simply connected space along that boundary kills the normal closure of the commutator of group elements, whenever the hypotheses of the Seifert-van Kampen theorem hold.

In the quotient space description of the Möbius band, follow $v=+1$ once along the rectangle. The seam sends its endpoint to the start of $v=-1$, which must also be traversed before returning to the original point. Under the deformation retraction onto the core circle of a Möbius band, these two stretches each travel once around the same circle. Orienting both circles by increasing longitudinal parameter gives a map of topological degree two. Equivalently, the group homomorphism on fundamental groups induced by inclusion of the boundary sends a chosen generator to twice the core generator, or to $c^2$ in multiplicative notation. Reversing the chosen boundary generator changes the sign, not the fact that it winds twice.

For a good pair $(X,A)$ with $A\ne \varnothing$, the quotient map induces natural isomorphismsHere $X/A$ collapses the entire subspace $A$ to one base point. A neighborhood deformation retraction and excision compare the relative homology with the quotient, so computing a pair becomes computing the reduced homology of one space.

If $A=\varnothing$, use the based convention $X/\varnothing =X_{+}$, adjoining a disjoint base point, to retain the formula. With the unbased quotient convention $X/\varnothing =X$, the formula would fail in degree zero. For the closed disk and its boundary sphere, the quotient is $S^n$ and recovers the relative homology of a simplex and its boundary.

The core circle is the image of the centre line in the rectangle defining a Möbius band. Its endpoints are identified, so it is a circle. The formuladefines a deformation retraction onto that circle: it respects $(0,v)\sim (1,-v)$, starts at the identity map, and fixes the core throughout. Consequently the Möbius band has fundamental group isomorphic to $\mathbb{Z}$, just like a circle; this alone cannot distinguish it from an annulus.

A deformation retraction of $X$ onto a subspace $A$ is a homotopy $H:X\times [0,1]\to X$ such that $H(x,0)=x$, $H(x,1)=r(x)\in A$, and $H(a,t)=a$ for every $a\in A$ and every $t$. Thus the retract is fixed pointwise throughout. With inclusion $i:A\hookrightarrow X$, the equations $r\circ i={\mathrm{id}}_A$ and $i\circ r\simeq {\mathrm{id}}_X$ exhibit a homotopy equivalence. See Hatcher, Algebraic Topology, Chapter 0, pp. 2–3.

Take the closed annulusAn exact deformation retraction onto its inner circle isThe denominator is nonzero everywhere on the annulus. If the starting radius is $\rho =\Vert x\Vert$, its radius at time $t$ is $(1-t)\rho +t$. This lies between $1$ and $\rho \le 2$, verifying that the entire homotopy stays in the original annulus. When $\rho =1$, the point is stationary, as required.

For $x=(0,2)$, times $0,1/2,1$ give $(0,2),(0,3/2),(0,1)$. A radius-$3/2$ loop similarly becomes a radius-$5/4$ loop halfway through, then the unit circle. Its angle and winding number are preserved. This is useful because computing the fundamental group of the annulus reduces to computing that of the circle; the explicit homotopy equivalence establishes why the answer transfers.

For a real vector bundle $E\to B$ with a fiber metric, the disk bundle isIts fiberwise boundary is the sphere bundle $S(E)=\{\Vert v\Vert =1\}$. The radial homotopy $H_t(v)=(1-t)v$ gives a deformation retraction onto the zero section, so $D(E)$ is homotopy equivalent to $B$.

If $B$ is a smooth manifold without boundary and the rank is positive, the total-space boundary is exactly $S(E)$. When $B$ is a manifold with boundary, the base boundary also contributes: the boundary faces are $S(E)$ and $D(E){|}_{\partial B}$, meeting along $S(E){|}_{\partial B}$. For the trivial vector bundle of rank one over $[0,1]$, the disk bundle is the rectangle $[0,1]\times [-1,1]$. Its top and bottom sides form the sphere bundle; its left and right sides come from the base boundary. The intersections are corners.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

The fundamental group ${\pi }_1(X,x_0)$ consists of based loops modulo based homotopy. Multiplication follows one based loop and then the other; the constant based loop represents the identity, and path reversal gives the inverse. Passing to homotopy classes makes this operation associative, as explained by well-defined loop concatenation.

Identify the circle with $S^1=\{z\in \mathbb{C}:|z|=1\}$ and choose basepoint $1$. Its based loopshave winding number $n$: positive $n$ counts counterclockwise turns, negative $n$ counts clockwise turns, and $n=0$ gives the constant based loop. Every based loop on the circle is joined by a based homotopy to exactly one ${\gamma }_n$. ConsequentlyThis is Hatcher's Theorem 1.7 in Algebraic Topology, §1.1, pp. 29–30. For example, two counterclockwise turns followed by one clockwise turn represent $2+(-1)=1$, rather than three turns. The winding graph in the figure above makes the signed arithmetic visible.

A proof outline uses the covering map $q:\mathbb{R}\to S^1$, $q(u)=e^{2\pi iu}$. By the path lifting theorem, a based loop has a unique lift starting at zero and ending at an integer $n$. Interpolating its lift to $s\mapsto ns$ produces a based homotopy to ${\gamma }_n$. The homotopy lifting theorem for a covering map keeps that endpoint unchanged during a based homotopy, proving uniqueness.

In the filled disk, every based loop $\gamma$ at $x_0$ has a based homotopy to the constant based loop:Both endpoints remain $x_0$, and the convex set property of the disk keeps every intermediate point inside. Thus ${\pi }_1(D^2,x_0)$ is the trivial group. Together with the deformation retraction above, this computes three concrete answers: the filled disk has trivial fundamental group, while the circle and the annulus have infinite cyclic groups.

Maps have computable algebraic shadows too. The continuous map $p_n:S^1\to S^1$, $p_n(z)=z^n$, fixes $1$ and sends ${\gamma }_k$ to ${\gamma }_{nk}$. Its induced group homomorphism is therefore $k\mapsto nk$, and its mapping degree is $n$. For $n=2$ the image consists of even integers, so this map cannot be a homotopy equivalence. Composing $p_2$ with $p_{-3}$ gives $p_{-6}$: map composition multiplies degrees, whereas concatenating based loops adds winding numbers. Distinct degrees distinguish these maps up to homotopy; this calculation does not classify arbitrary topological spaces.

Historically, Poincaré's 1895 Analysis Situs introduced the fundamental group; its original §12 explicitly names it. The CNRS transcription of Poincaré's §12, Groupe fondamental provides a direct view of this early formulation.

Removing an embedded open disk from the torus leaves a topological surface with one boundary circle. In the square model, radial expansion away from the removed disk gives a deformation retraction onto the square boundary with opposite sides identified, a wedge of two circles. Its fundamental group is therefore the free group on two group generators.

Attach a closed interval to a sphere $S^2$ at two distinct endpoints, with its interior disjoint from the sphere. The fundamental group is infinite cyclic group. Remove a point of the sphere away from the endpoints: the remaining plane has a deformation retraction onto an arc between the endpoints, keeping the added closed interval fixed. This gives a circle. An open cap about the removed point is simply connected, and its punctured overlap has trivial map into the first fundamental group because it lies in the plane. The Seifert-van Kampen theorem therefore adds no further relator.

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]

Homotopy theory studies topological spaces and continuous maps through continuous deformation. A homotopy from $f:X\to Y$ to $g:X\to Y$ is a continuous map $H:X\times [0,1]\to Y$ with $H(x,0)=f(x)$ and $H(x,1)=g(x)$. Replacing a space by a homotopy equivalent one preserves its deformation-level information, including its homotopy groups.

For example, the punctured plane retracts onto its unit circle throughThe coefficient stays positive, so the deformation never crosses the removed origin, and points already on the unit circle remain fixed. This is a deformation retraction.

The same viewpoint studies loops, higher-dimensional spheres and maps between spaces. It retains obstructions that survive continuous deformation while allowing distances, angles and the thickness of a region to change. A homotopy equivalence is weaker than a homeomorphism.

For $C_2\ast \mathbb{Z}\to S_3$ taking the two generators to a transposition and a $3$-cycle, the kernel's six-sheeted regular covering space consists of three spheres and two circles. Each sphere has two marked points, and each circle meets one point on each sphere. It retracts onto a rank-four graph by replacing each sphere by a meridian arc, but admits no deformation retraction onto any graph because its degree-two homology is ${\mathbb{Z}}^3$.

The Möbius band is the topological surface obtained by gluing the ends of a rectangle with a half-turn. Precisely, it is the quotient spaceThe twist identifies the two apparent long edges into one boundary circle. Following the core circle of a Möbius band reverses local handedness, so this surface is not orientable. An untwisted untwisted band has two boundary circles and is orientable: the distinction survives stretching and bending because orientability is a topological invariant.

To explore the construction with paper, mark the two long edges in different colours, join the short ends after a half-turn, and trace an edge. The trace passes from one marked edge to the other without ever leaving the boundary. Separately, trace the centre line and track a local transverse arrow: it comes back reversed after one circuit. The boundary closes after twice the travel around the centre, while the transverse arrow needs two circuits to return to its original direction.

This original mathematical model has central radius $R=2$ and half-width $a=0.55$:Its seam obeys $F(2\pi ,w)=F(0,-w)$. The dotted circle is the core circle of a Möbius band; the teal curve follows the single boundary through $0\le \theta \le 4\pi$. At the same point on the core, the arrows for $\theta =0$ and $\theta =2\pi$ are opposite. The graph compares a continuously followed unit normal with its starting value in the ambient space; the endpoint reversal, rather than any intermediate zero of the graph, establishes the obstruction. It is not a statement about parallel transport.

The Möbius band has the homotopy type of a circle: a deformation retraction shrinks each width coordinate to zero. Nevertheless its boundary represents twice a generator of its fundamental group. The boundary of a Möbius band winds twice around its core, so with compatible generator choices the boundary generator maps to $c^2$ in ${\pi }_1(M)\cong \mathbb{Z}$.

The historical account by Julyan H. E. Cartwright and Diego L. González places the mathematical discovery by August Möbius and Johann Listing in 1858, while investigating representations of twisted bands predating that work. A familiar-looking picture and a recognition of the underlying topological invariant are different kinds of evidence. ^[ref]

The total space of the orientation line bundle is orientable even when its base is not: an orientation reversal in a base coordinate transition is matched by a reversal in the line coordinate, so their product has positive sign. Fiberwise multiplication by $1-t$ gives a deformation retraction onto the zero section. For the real projective plane this total space is $(S^2\times \mathbb{R})/((v,t)\sim (-v,-t))$, an oriented noncompact three-manifold homotopy equivalent to the plane.

A rank-$k$ complex vector bundle is a smooth map $\pi :E\to B$ whose fibers are $k$-dimensional complex vector spaces, with local diffeomorphisms ${\Phi }_i:{\pi }^{-1}(U_i)\to U_i\times {\mathbb{C}}^k$ commuting with projection and linear on every fiber. Define transition functions of a vector bundle byThey are smooth and obeyon the appropriate double/triple overlaps. The structure group of a vector bundle is a Lie subgroup $G\subset GL(k,\mathbb{C})$ in which a chosen set of transitions takes values. It records allowed changes of fiber frame; different trivializations may exhibit different reductions of that group.

Using the same cocycle, glue the disjoint union of $U_i\times G$ byThe cocycle makes this an equivalence relation and makes its local charts compatible. Right multiplication $(b,h{)}_i\cdot a=(b,ha{)}_i$ is well-defined because it commutes with the left gluing matrices. It is free and transitive on each fiber. These charts therefore construct the associated right principal bundle $P\to B$ with group $G$.

For the natural line realization, take the complex tautological line bundle $L=\{(\ell ,v):\ell \in {\mathbb{CP}}^1,v\in \ell \subset {\mathbb{C}}^2\}$. Its nonzero vectors identify with ${\mathbb{C}}^2\setminus \{0\}$ by $(\ell ,v)\mapsto v$, with projection $v\mapsto [v]$. On the two standard charts, use $\zeta =z_2/z_1$ and $\eta =z_1/z_2=1/\zeta$. The local frames are $e_0=(1,\zeta )$ and $e_1=(\eta ,1)$, so $e_1=e_0/\zeta$. Normalize them using the ambient Hermitian norm:If $v=t_0{u}_0=t_1{u}_1$, then $t_0=(|\zeta |/\zeta )t_1$. Thus the unitary transitions of the tautological line over the projective line areThese are smooth on the overlap $\zeta \ne 0$. A Hermitian metric on a smooth complex vector bundle generally permits precisely this unitary reduction; here the ambient norm supplies it explicitly.

The corresponding principal circle bundle consists of unit vectors in the tautological line. Sending $(\ell ,v)$ to $v$ identifies its total space with $S^3\subset {\mathbb{C}}^2$, and the right circle action is $v\cdot e^{i\theta }=e^{i\theta }v$. Identify ${\mathbb{CP}}^1$ with $S^2$ usingIts affine-coordinate inverse on the chart $Z\ne -1$ is $\zeta =(X-iY)/(1+Z)$; the other chart handles the missing pole. This gives a diffeomorphism, and the projection becomes the Hopf fibration

There is a distinction if the stated punctured-bundle identification is interpreted only as a projection-preserving diffeomorphism, rather than a fiber-linear identification. It does not determine whether the complex line is tautological or dual: for a nonzero $\varphi \in L_{\ell }^{\ast }$, the unique $v\in \ell$ with $\varphi (v)=1$ gives a smooth map $L^{\ast }\setminus 0\to L\setminus 0$. Locally it is inversion of a nonzero complex number. Thus punctured line bundles do not determine their duality sign.

For completeness, these are the only possible signs here. Complex line bundles over $S^2$ are described by the clutching construction with a loop in ${\mathbb{C}}^{\ast }$, reducible to $S^1$, and its integer winding number determines the bundle: a zero-degree ratio has a logarithm and can be removed by changes of trivializations extending over the two disks. For winding number $d$, the unit circle bundle has fundamental group $\mathbb{Z}/d\mathbb{Z}$, as the two solid-torus trivializations identify the common fiber generator and impose its $d$th power as the equatorial gluing relation. There is a fiberwise radial deformation retraction from the punctured line bundle onto its unit circle bundle. Since ${\mathbb{C}}^2\setminus 0$ retracts to simply connected $S^3$, the hypothesis forces $|d|=1$.

Consequently the allowed unitary transitions are the displayed $g_{01}$ or its reciprocal, according to the bundle's sign. For the dual, the same underlying Hopf projection has the inverse principal action $v\cdot e^{i\theta }=e^{-i\theta }v$. In either case the principal total space and projection are $S^3\to S^2$; the purely diffeomorphic hypothesis alone leaves the circle-action sign unspecified.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.

The Excision theorem for singular homology states that if $Z\subseteq A\subseteq X$ and $\overline{Z}\subseteq {\mathrm{int}}_{X}A$, inclusion induces isomorphisms $H_q(X\setminus Z,A\setminus Z;\mathbb{Z})\cong H_q(X,A;\mathbb{Z})$ for all $q$. For an open cover $X=U\cup V$, its small-chain form says that the subcomplex of singular chains whose individual simplices lie in $U$ or $V$ computes $H_{\ast }(X)$. Concretely, repeated barycentric subdivision puts each finite chain in that subcomplex, and subdivision is chain-homotopic to the identity. There is then a short exact sequence of chain complexesIts long exact sequence in homology, with excision identifying the last complex's homology with $H_{\ast }(X)$, is the Mayer–Vietoris sequence:

The Real projective space ${\mathbb{RP}}^1$ is a circle. The projective plane has one cell in each dimension $0,1,2$, with cellular homology boundary maps $d_2=2$ and $d_1=0$. Thus, writing $A_q=H_q({\mathbb{RP}}^2;\mathbb{Z})$, we have $A_0=\mathbb{Z}$, $A_1=\mathbb{Z}/2$ and $A_q=0$ for $q\ge 2$. Cover the circle by two open arcs with a two-component intersection and multiply this cover by the projective plane. In the Mayer–Vietoris sequence the overlap-to-cover map isIts kernel and cokernel are each $A_q$. Consequently $0\to A_q\to H_q(X)\to A_{q-1}\to 0$ for $q\ge 1$. In degree one the quotient is free, so this sequence splits; in degree two it identifies the group directly with $A_1$. The integral homology of a circle times a real projective plane is therefore

Without a compactness restriction, the answer to the final existence question is yes: there is a noncompact oriented four-manifold. Explicitly, takeThe involution is free and reverses orientation on both $S^2$ and $\mathbb{R}$, hence preserves their product orientation. Its quotient is consequently an oriented three-manifold, the orientable total space of the orientation line bundle of the projective plane. The contraction $[v,t]\mapsto [v,(1-s)t]$ is a deformation retraction onto ${\mathbb{RP}}^2$, so $Y\simeq X$.

If a closed oriented manifold is intended, the answer instead is no. A closed connected oriented $d$-manifold has $H_d\cong \mathbb{Z}$ and no homology above $d$. Here the highest nonzero positive-dimensional homology is the torsion group $H_2=\mathbb{Z}/2$, so no dimension can supply that required top fundamental class. The distinction is necessary because the printed wording does not require the oriented manifold to be closed or three-dimensional.

Choose open neighborhoods of the sphere and circle in the wedge, each enlarged by a small contractible neighborhood of the attaching point in the other piece. They deformation retraction to $S^2$ and $S^1$ and have contractible intersection. Van Kampen therefore givesThe generator traverses the circle once; the sphere contributes no fundamental-group generator.