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If an embedded codimension-$k$ submanifold $X\subset M$ is $f^{-1}(y)$ for a regular value $y$ of a smooth map $f:M\to {\mathbb{R}}^k$, then $df$ identifies its normal bundle with the trivial bundle $X\times {\mathbb{R}}^k$. A submanifold with nontrivial normal bundle therefore cannot be such a regular level set. The core circle of the Möbius band is a codimension-one example.

Integral groups for $M_4$. Put $r=n-1$. The space $M_n$ is the inversion mapping torus of a torus, fibred over $S^1$ with fiber $T^r$. The cohomology ring of a torus is the exterior algebra on $r$ degree-one generators. Inversion acts as $-1$ on each such generator, hence as $(-1{)}^q$ on $H^q(T^r;\mathbb{Z})\cong {\mathbb{Z}}^{\left(\genfrac{}{}{0ex}{}{r}{q}\right)}$.

The Wang sequence therefore givesThe right-hand term is free, so this short exact sequence splits as a sequence of groups. In even fiber degree, $F^{\ast }-1=0$; in odd fiber degree it is multiplication by $-2$. For $r=3$ this computes the integral cohomology of an inversion mapping torus:The top torsion group is consistent with $M_4$ being nonorientable: inversion of its three-dimensional fiber reverses orientation.

Mod-two groups. Over ${\mathbb{F}}_2$, inversion acts as the identity on the fiber cohomology. The Wang sequence givesThuswith out-of-range binomial coefficients interpreted as zero. These are exactly the dimensions of the graded groups $H^{\ast }(T^n;{\mathbb{F}}_2)$.

The intersection pairing for $M_2$. The mapping torus of reflection of the circle is the Klein bottle. Let $a$ be the section loop through a fixed point of the reflection and $b$ a fiber circle. They form a basis of $H_1(M_2;{\mathbb{F}}_2)$. The section has a normal neighborhood homeomorphic to a Möbius band, so it is one-sided and a transverse displacement meets it once modulo two. The fiber is two-sided and can be displaced disjointly. The two loops meet once. The mod-two intersection pairing of the Klein bottle therefore has matrixTake the evaluation-dual basis $t,x\in H^1(M_2;{\mathbb{F}}_2)$, with $t(a)=1,t(b)=0$ and $x(a)=0,x(b)=1$. By Poincare duality, the cup product matrix in this dual basis is $Q^{-1}$, not $Q$:Writing $w$ for the nonzero top class gives $t^2=0$, $tx=w$ and $x^2=w$. Hence the cohomology ring isThese relations already force all degrees above two to vanish.

The ring for general $n$. For each of the $r$ fiber coordinates, projection induces $M_n\to M_2$. Pull back the classes above, calling the common base class $t$ and the fiber-coordinate classes $x_1,\dots ,x_r$. Naturality of the cup product gives $t^2=0$ and $x_i^2=t{x}_i$. The square-free products $x_I={\prod }_{i\in I}{x}_i$ restrict to the exterior algebra basis of the fiber cohomology. The Leray-Hirsch theorem then says that $x_I$ and $t{x}_I$ form an additive basis. This proves the full mod-two cohomology ring of an inversion mapping torus:For $n\ge 2$, $x_i^2=t{x}_i$ is nonzero by this basis description. In the cohomology ring of a torus every degree-one class squares to zero: the generators square to zero and the cross terms occur twice in characteristic two. The degree-one cup-square obstruction to ring isomorphism therefore proves that the rings are not isomorphic for any $n\ge 2$, despite their isomorphic graded groups. For $n=1$, both spaces are circles and the rings are isomorphic. Even if grading is forgotten, the rings differ for $n\ge 2$: every element of the torus ring has square either zero or one, whereas $x_i^2=t{x}_i$ is nonzero and is not the unit.

Integral cup products in $M_4$. To specify $H^{\ast }(M_4;\mathbb{Z})$ also as a ring, let $a\in H^1$ be the pullback of the positive generator of $H^1(S^1;\mathbb{Z})$. Let $\rho$ denote reduction modulo two and set ${\tau }_i=\hat{\beta }(x_i)$ using the integral Bockstein homomorphism. The degree-one Bockstein square identity gives $\rho {\tau }_i=x_i^2=t{x}_i$, so the ${\tau }_i$ are the three independent order-two classes in degree two. Choose free classes $b_{12},b_{13},b_{23}\in H^2$ which restrict to the corresponding two-fold fiber products and have reductions $\rho b_{ij}=x_i{x}_j$. Such choices exist: reduction in degree two is surjective because $H^3(M_4;\mathbb{Z})$ is free, and adding the ${\tau }_i$ removes any $t{x}_i$ terms from an initial lift.

The Wang sequence identifies $a{b}_{12},a{b}_{13},a{b}_{23}$ as a free basis of $H^3$. Let $\kappa$ generate $H^4\cong \mathbb{Z}/2$, with $\rho \kappa =t{x}_1{x}_2{x}_3$. Reduction in degree four is an isomorphism. The integral cup products in the four-dimensional inversion mapping torus are consequently determined byTogether with the unit, graded commutativity and vanishing above degree four, these give every product. For example, $\rho (b_{12}{b}_{13})=x_1^2{x}_2{x}_3=t{x}_1{x}_2{x}_3$, whereas $\rho (b_{12}^2)=t^2{x}_1{x}_2=0$. The products $a{\tau }_i$ vanish because they are torsion in the free group $H^3$.

A useful closed-cover version of the Seifert-van Kampen theorem is the following. Let $X=A\cup B$ with $A,B,A\cap B$ path-connected closed subspaces and basepoint in their intersection. Assume there are open neighbourhoods $U,V$ covering $X$ that deformation retract onto $A,B$, while $U\cap V$ deformation retracts onto $A\cap B$, with the induced inclusions compatible up to based homotopy. ThenThe neighbourhood hypothesis, available for the present finite subcomplex gluing, matters: arbitrary closed covers need not satisfy this theorem.

A Möbius band retracts onto its core circle, with generator $m$; its boundary represents $m^2$. The torus has generators $u,v$ and relation $[u,v]=1$. The attaching circle is the $u$ circle. Reversing the attaching homeomorphism merely replaces $u$ by its inverse and changes no isomorphism type. The closed-cover theorem givesMapping $m\mapsto 0$, $v\mapsto 1$ surjects onto $\mathbb{Z}$, so the group is infinite. Imposing the extra relation $m^2=1$ gives the quotient $C_2\ast \mathbb{Z}$, a nonabelian free product, since alternating reduced words $mv$ and $vm$ differ. A quotient of an abelian group would be abelian, so the fundamental group is nonabelian.

An immersed submanifold of $M$ is a manifold $Y$ equipped with an injective immersion $\psi :Y\to M$. It is an embedded submanifold when $\psi$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding. For irrational $\alpha$, the irrational winding of the torusis an injective immersion $\mathbb{R}\to T^2$ with dense image, and therefore is not an embedding. If $Y$ is compact, however, an injective immersion is a continuous bijection from a compact space to its image in the Hausdorff manifold $M$; its inverse is continuous. Thus the compact injective immersion is an embedding theorem makes $\psi (Y)$ embedded.

Now let $X\subset M$ be embedded, with $\dim M=m$, $\dim X=m-k$, and $p\in X$. Apply the constant rank theorem to its inclusion. After choosing coordinates and reordering them, there is a neighborhood $U$ of $p$ with coordinates $(x^1,\dots ,x^m)$ for whichThis is the slice chart for an embedded submanifold.

It is false that every embedded submanifold is the inverse image of a regular value of a map to a Euclidean space. If a codimension-$k$ submanifold is $f^{-1}(y)$ for a regular value of $f:M\to {\mathbb{R}}^k$, the differentials of the component functions give a global frame of its conormal bundle, so its normal bundle is trivial. The core circle of the Möbius band is embedded but has the nontrivial Möbius normal line bundle. This is the normal-bundle obstruction to being a regular level set.

Finally, an inductive spinning construction gives the requested torus. Place an embedding $F:N^m\hookrightarrow {\mathbb{R}}^{m+1}$ in the half-space $F_{m+1}>0$ and defineThe positive radius makes this map injective, and its differential is injective in both the $N$ and circle directions; compactness then makes it an embedding. Starting with $S^1\hookrightarrow {\mathbb{R}}^2$ and iterating proves the embedding of the n-dimensional torus in codimension one:

The zero section $C=\{t=0\}\subset M$ is a circle. Its normal bundle is the real line bundle obtained fromso its clutching map reverses sign once around $C$. Its mod-two Euler class of a real line bundle, equivalently its first Stiefel–Whitney class, therefore satisfiesIf $M$ were orientable, the splittingand the orientation of the circle would orient ${\nu }_C$, forcing this mod-two Euler number to vanish. This contradiction proves that the Möbius band is nonorientable.

Yes. For any $R>1$, the formuladefines the standard smooth embedding of the open Möbius band into ${\mathbb{R}}^3$. Replacing $(x,t)$ by $(x+2\pi ,-t)$ leaves the displayed point unchanged. The image is the interior of a compact Möbius strip, so the embedding fails to be a proper map; this is why it does not contradict part (d).

Cut the octagon for $S_3$ along a diagonal joining the appropriate two vertex classes. The two resulting polygons can be reattached along that diagonal in the opposite order. Reading the new boundary word gives the square wordwhich is exactly the edge identification displayed for $S_4$. The cut-and-paste map is affine on the two pieces and respects every paired edge, so it descends to a homeomorphism $S_3\cong S_4$. Both are the Klein bottle.

After deleting an open disc, the punctured Klein bottle can be realized as a boundary connected sum of two embedded Möbius strips, and hence embeds in ${\mathbb{R}}^3$. The closed Klein bottle cannot embed: every connected closed surface embedded in ${\mathbb{R}}^3$ is two-sided and therefore orientable, by the Jordan-Brouwer separation theorem, whereas the Klein bottle is nonorientable. Thus

Cut the square model of the Klein bottle along the midline parallel to the pair of sides whose identification reverses orientation. Each half becomes a Möbius band, and the two new boundary circles are identified. If $a$ and $b$ are the core loops of these two bands, each boundary travels twice around its core. The Seifert-van Kampen theorem therefore gives

Writing $x=a^{-1}b$ and $y=a$ converts this into the standard Klein-bottle presentationThe orientation homomorphism sends both $a$ and $b$ to the nonzero element of $C_2$. Its kernel is the index-two subgroupBy the classification of connected covering spaces, $H$ determines the orientation double cover of the Klein bottleChoosing torus generators $\alpha ,\beta$ along the two translation directions, we may takeThese commute: the standard relation says that $a$ conjugates $a^{-1}b=x$ to $x^{-1}$, so $a^2$ centralizes $x$.

The space $Y$ is the mapping cylinder of $p$ with its unused end omitted. Sliding $(x,t)$ to $(x,0)$ gives a deformation retraction of $Y$ onto the bottom quotient $T^2/\sim \cong K$. Hence

Finally suppose $U\subseteq X$ is open and homeomorphic to $Y$, and identify the bottom copy of $K$ inside $U$. This $K$ is compact and therefore closed in the Hausdorff space $X$. Apply van Kampen, in its groupoid form if intersections are disconnected, toThe intersection $U\setminus K$ deformation retracts to a positive-height torus, and its map to ${\pi }_1(U)\cong {\pi }_1(K)$ has image preciselyThe orientation homomorphismis zero on this intersection, so it is compatible with the trivial homomorphism from ${\pi }_1(X\setminus K)$ to $C_2$. The pushout property in van Kampen extends it to a surjectionThereforeThis is the Klein-bottle mapping-cylinder obstruction to simple connectivity.

Given group homomorphisms $i_A:C\to A$ and $i_B:C\to B$, their amalgamated free product is a group $A{\ast }_{C}B$ with maps $j_A:A\to A{\ast }_{C}B$ and $j_B:B\to A{\ast }_{C}B$ satisfying $j_A{i}_A=j_B{i}_B$, such that every pair $f_A:A\to H$, $f_B:B\to H$ satisfying $f_A{i}_A=f_B{i}_B$ factors through a unique homomorphism $A{\ast }_{C}B\to H$.

The Seifert-van Kampen theorem says that if $X=U_1\cup U_2$, where $U_1,U_2$, and $U_1\cap U_2$ are open and path-connected and $x_0\in U_1\cap U_2$, then the inclusion maps induce

Here is a direct proof of the requested generation statement. For a based loop $\gamma$ in $X$, the Lebesgue number lemma supplies a subdivisionsuch that every arc $\gamma {|}_{[t_{j-1},t_j]}$ lies in one $U_{{\epsilon }_j}$. Combine adjacent arcs assigned to the same set, so each transition point $z_j=\gamma (t_j)$ lies in $U_1\cap U_2$. Since the intersection is path-connected, choose a path $q_j$ in it from $x_0$ to $z_j$, taking $q_0=q_m$ constant. Thenis a loop in $U_{{\epsilon }_j}$. In the product of their classes, each $\overline{q_j}\ast q_j$ cancels by path reversal, leaving $[\gamma ]$. Thus the two inclusion images generate ${\pi }_1(X,x_0)$, as summarized by generation of a fundamental group by two open sets.

Now let $a,b$ be the standard generators in the fundamental group of the torus. A Möbius band retracts to a core circle, while its boundary traverses that core twice. If $x$ and $y$ denote the core classes of the two attached bands, the Seifert-van Kampen theorem gives

The two attaching classes $A=ab$ and $B=a^2{b}^3$ commute. Their exponent matrix has determinantso they form a basis of $⟨a,b⟩\cong {\mathbb{Z}}^2$, withSubstituting $A=x^2$ and $B=y^2$ eliminates $a,b$ and leaves the two-generator one-relator presentationThis is the calculation in two Möbius bands attached to a torus.

Finally send $x$ to $(12)$ and $y$ to $(23)$ in the symmetric group $S_3$. Both squares are the identity, so the relator is satisfied, and these transpositions generate $S_3$. We obtain a surjective group homomorphism from ${\pi }_1(Y)$ onto the nonabelian group $S_3$. Therefore ${\pi }_1(Y)$ is nonabelian.