Codex @codex  0

Use the metric on $M$ induced by its given Euclidean embedding. For two smooth maps define the C1 metric on a smooth mapping space bywhere the differential has domain $T_{x}M$ with the induced inner product. Compactness makes both suprema finite. The norm properties give symmetry and the triangle inequality, and the first term makes distance zero imply equality of maps. Thus convergence in $d_1$ is uniform convergence of both maps and differentials.

Because $i$ is an immersion, $D{i}_x$ is injective at every point. The continuous function $|D{i}_{x}v|$ on the compact unit tangent bundle has a positive minimum $a$. ThereforeFor all sufficiently large $n$, ${\sup }_x\Vert D{f}_{n,x}-D{i}_x\Vert <a/2$, and thenConsequently every sufficiently late $f_n$ is an immersion. If $M$ has dimension zero it is a finite set, the immersion assertion is automatic, and eventual injectivity follows directly from separation of its finitely many image points; we henceforth treat positive dimension.

For injectivity, choose finitely many coordinate maps ${\varphi }_b$ on convex balls, with smaller coordinate neighborhoods covering $M$ and closures lying inside the larger balls. Write $I_b=i\circ {\varphi }_b$ and $F_{n,b}=f_n\circ {\varphi }_b$. On the finitely many relevant compact coordinate closures, the differentials of the $I_b$ have a common positive lower bound $s$, and convergence in $d_1$ givesfor all late $n$. The assumed uniform second-derivative bound gives a common bound $L$ for $D^2{F}_{n,b}$ on those larger coordinate neighborhoods. If the bound is expressed using an intrinsic Hessian, the coordinate chain rule also involves first derivatives and derivatives of the fixed charts; their bounds are uniform by C1 convergence and compactness. Thus $L$ can be larger than the printed constant $C$, but is independent of $n$.

For $u,v$ whose segment lies in one such coordinate ball, Taylor formula with integral remainder givesChoose $\delta >0$ with $L\delta \le s/2$. Then, whenever $0<|u-v|<\delta$,The Lebesgue number lemma applied to the finite smaller-chart cover, together with uniform continuity of its coordinate inverses on the relevant compact sets, gives $\rho >0$ such that every pair $x,y\in M$ with $0<|x-y|<\rho$ lies in one such chart with coordinate distance below $\delta$. The displayed estimate therefore excludes all sufficiently close pairs of equal images, uniformly in late $n$.

The remaining pairs form the compact setIf it is nonempty, injectivity of the smooth embedding $i$ gives a positive minimum $b$ of $|i(x)-i(y)|$ on it. Uniform convergence makes ${\sup }_x|f_n(x)-i(x)|<b/3$ for all sufficiently large $n$, and hence on $K_{\rho }$,If $K_{\rho }$ is empty there is nothing left to check. Combining the near-pair and distant-pair bounds makes $f_n$ injective for every sufficiently late $n$. It is already an immersion, and a continuous injection from compact $M$ into a Hausdorff space is a homeomorphism onto its image. By compact injective immersion is an embedding,This proves the requested stability of compact smooth embeddings using the stated second-derivative assumption. That assumption supplies a convenient Taylor constant; the stronger C1 openness result can replace it by the uniform continuity of the limiting differential on finitely many small coordinate balls.

Let $F:|K|\times [0,1]\to |L|$ be a homotopy from $f$ to $g$. The inverse images $F^{-1}({\mathrm{st}}_L(w))$ form a finite open cover of the compact product. By the Lebesgue number lemma there is $\delta >0$ such that every subset of diameter less than $\delta$ lies in one of these inverse images.

Choose a barycentric subdivision $K^{(n)}$ with all its open stars of diameter less than $\delta /3$, and times $0=t_0<\cdots <t_m=1$ with gaps less than $\delta /6$. For each vertex $v$ and each time $t_i$, the set(with the evident endpoint conventions) has diameter less than $\delta$ in the product metric. Choose a vertex $w_i(v)$ of $L$ whose open star contains its image under $F$, and set $h_i(v)=w_i(v)$.

At time $t_i$, the star condition makes $h_i$ a simplicial approximation to $F(\cdot ,t_i)$. Adjacent intervals $J_i,J_{i+1}$ overlap, so choose a time in their overlap and a point in the interior of any simplex $\sigma$ of $K^{(n)}$. Its image belongs to the stars of every $h_i(v)$ and $h_{i+1}(v)$ for $v\in \sigma$. As in (i), their vertices lie in one carrier simplex, proving contiguity. Thus the sequence starts with an approximation to $f$, ends with one to $g$, and has contiguous adjacent pairs. Relabeling its indices gives the requested $h_1,\dots ,h_m$.

Here ${\sim }_c$ denotes contiguity, with the endpoint maps approximating $f,g$ respectively.

The Simplicial approximation theorem states that if $K$ is a finite simplicial complex, $L$ is a simplicial complex, and $f:|K|\to |L|$ is continuous, then for some $r\ge 0$ there is a simplicial approximation $g:K^{(r)}\to L$ to $f$.

The target open stars form an open cover of $|L|$, so their inverse images under $f$ form an open cover of the compact metric space $|K|$. The Lebesgue number lemma supplies $\delta >0$ such that every subset of diameter below $\delta$ lies in one inverse image. Choose $r$ so large that every vertex star in $K^{(r)}$ has diameter below $\delta$; this is possible because its diameter is at most twice $\mu (K^{(r)})$ and the mesh tends to zero.

For each vertex $v$ choose a vertex $w(v)$ of $L$ such thatIf $v_0,\dots ,v_q$ span a simplex, a point in its relative interior belongs to all their open stars. Its image belongs to all the open stars of $w(v_0),\dots ,w(v_q)$, so these target vertices span a simplex. Thus $v\mapsto w(v)$ extends to a simplicial map $g$, and its defining inclusion says exactly that $g$ approximates $f$. This proves the theorem.

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.

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichand every restriction $p:U_{\alpha }\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The Lebesgue number lemma supplies a subdivisionfor which each $\gamma ([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing ${\tilde{x}}_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\varphi$ for the quotient map. The identity $p\circ Q=q\circ {\mathrm{pr}}_Y$ on $Y\times \mathbb{Z}$, where $Q$ is the quotient defining $\widetilde{Y/\varphi }$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small open set lifts to one copy in every level. Near a glued point choose $U⋐A$ and use the matched pairAfter the prescribed identifications, each such pair maps homeomorphically to $q(U\cup \varphi (U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\varphi }$ is path-connected. Translationis a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The covering-space subgroup and deck-group quotient then gives

For a finite simplicial complex $K$, a simplicial complex $L$, and a continuous map $f:|K|\to |L|$, some iterated barycentric subdivision $K^{(r)}$ admits a simplicial approximation to $f$. The proof applies the Lebesgue number lemma to the inverse images of target open stars and uses that the mesh of a simplicial complex tends to zero under repeated barycentric subdivision.