Machine-translated from Chinese.
浸入, 淹没, 嵌入
浸入, 淹没有着典范表示. 局部常秩映射也有类似的典范表示.
前两者证明方法是标准化$D\widehat f,$ 然后补全维度构造局部微分同胚.
利用这一微分同胚改造坐标系即可.
浸入+同胚=嵌入, 即嵌入子流形的子空间拓扑和诱导拓扑一致.
由浸入的局部表示(与同胚性质), 子流形具有子流形坐标卡, 为切片.
正则值
对于光滑映射$f:M\rightarrow N,$ 若其在$p\in M$为淹没, 则$p$为正则点,
否则为临界点. 临界点的像称为临界值.
$N$上不是临界值的点称为正则值. 也就是说,
正则值的原像都要是正则点(淹没), 允许为空集.
定理 1 (正则值原像定理). 设$f:M\rightarrow N$为光滑映射, $q\in N$为$f$的正则值. 那么$f^{-1}(q)$为$(\dim M-\dim N)$维嵌入子流形.
$\,\forall\,p\in f^{-1}(q),$ 由淹没的局部表示,
有坐标邻域$(U,\varphi)\ni p,$ $(V,\psi)\ni q,$ 使得
$\widehat f:\varphi(U)\rightarrow \psi(V)$ 为投影映射, $\varphi(p)=0,$
$\psi(q)=0$. 于是
$$
\widehat f^{-1}(0)=\{(0,y)|y\in \mathbb{R}^{\dim M-\dim N}\}=\varphi\circ f^{-1}(q).
$$
这就给出了$f^{-1}(q)$作为$(\dim M-\dim N)$维切片的子流形坐标卡$(U\cap f^{-1}(q),{\color{red}\pi_y}\circ \varphi)$.
定理 2 (唱片引理). 设$M,N$为维数相同的光滑流形, $M$紧致. 设$f:M\rightarrow N$为光滑映射. 若$q\in N$为$f$的正则值, 那么$f^{-1}(q)\subset M$为有限点集$\{p_i\}_{i=1}^k$, 且存在$q$在$N$中的开邻域$V,$ 使得$f^{-1}(V)=\bigsqcup_{i=1}^k U_i,$ $p_i\in U_i,$ 且$f|_{U_i}$为微分同胚.
由正则值原像定理, $f^{-1}(q)=\{p_i\}$为 $0$ 维子流形,
有子流形坐标邻域$W_i,$ 两两无交. 由于$M$紧致,
若$\{p_i\}$为无限点集将有聚点, 而聚点处取不到这样的坐标邻域$W_i$.
或$f^{-1}(q)$为闭集, 故紧.
由于$f$既是浸入又是淹没, 为局部微分同胚.
不妨设它在每个$W_i$上为微分同胚. 接下来取
$$
V=\bigcap_{i=1}^k f(W_i) \setminus f(M\setminus \bigcup_{i=1}^k W_i).
$$
其中$M\setminus \bigcup_{i=1}^k W_i$为闭集, 紧集中闭集是紧的.
从而$f(M\setminus \bigcup_{i=1}^k W_i)$为紧集, 进而为闭集.
于是$V$的确是开邻域, 且$f^{-1}(V)\subset \bigcup_{i=1}^k W_i.$
由于$f|_{W_i}$为微分同胚,
$\,\exists\,U_i{\color{red}{(=f^{-1}(V)\cap W_i)} }\subset W_i,$
使得$f^{-1}(V)=\bigsqcup_{i=1}^k U_i,$ 且$f|_{U_i}$为微分同胚.
定理 3 (Sard定理). 光滑映射$f:M\rightarrow N$的临界值集为零测集, 几乎处处为正则值(进而稠密).
横截
设$f:M\rightarrow N$为光滑映射, $Z\subset N$为嵌入子流形.
若$\,\forall\,p\in f^{-1}(Z),$
$$
df_p(T_pM)+T_{f(p)}Z=T_{f(p)}N,
$$
则称$f$与$Z$横截, 记为$f\pitchfork Z.$ 当$Z$为单点集时,
$Z$就是$f$的正则值.
特别地, 若$M\subset N$为嵌入子流形, 那么称$M\pitchfork Z$,
若放入映射$\iota:M\rightarrow N$与$Z$横截. 此时有
$$
T_pM+T_pZ=T_pN,\quad \,\forall\,p\in M\cap Z.
$$
类似地, 也可以定义两映射横截, 此时不见得要求映射为嵌入.
定理 4 (横截原像定理). 设$f:M\rightarrow N$为光滑映射, $Z\subset N$为嵌入子流形. 若$f\pitchfork Z,$ 则$f^{-1}(Z)$为$M$中的嵌入子流形, 且$\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$
由于$Z$为嵌入子流形, $\,\forall\,q\in Z,$ 有坐标卡$(V,\psi),$ $q\in V,$
使得$\psi(Z\cap V)\subset \mathbb{R}^{\dim Z}$为切片.
取投影$\pi:\mathbb{R}^{n}\rightarrow \mathbb{R}^{\operatorname{codim}_N Z},$
那么$\psi (Z\cap V)=\pi^{-1}(0).$
断言$g:=\pi \circ \psi \circ f:f^{-1}(V)\rightarrow \mathbb{R}^{\operatorname{codim}_NZ}$以$0$为正则值.
只需说明$\,\forall\,p\in f^{-1}(Z)\cap f^{-1}(V)=f^{-1}(Z\cap V),$
$dg_p$为淹没. 由横截性条件,
$$
d\psi \circ df_p(T_pM)+ T_{\psi\circ f(p)}\psi(Z)=T_{\psi\circ f(p)}\psi(N)=\mathbb{R}^n,\quad T_{\psi\circ f(p)}\psi(Z)=\mathbb{R}^{\dim Z}.
$$
再作用$d\pi=\pi,$ 即有
$$
dg_p(T_pM)=d\pi\circ d\psi\circ df_p(T_pM)=\mathbb{R}^{\operatorname{codim}Z}.
$$
由正则值原像定理,
${\color{red}{g^{-1}(0)=f^{-1}(Z\cap V)=f^{-1}(Z)\cap f^{-1}(V)} }$为嵌入子流形,
且
$$
\dim g^{-1}(0)=\dim M-\operatorname{codim}_NZ,\quad \operatorname{codim}_M (f^{-1}(Z)\cap f^{-1}(V))=\operatorname{codim}_NZ.
$$
由$q\in V$的任意性即有$f^{-1}(Z)$为嵌入子流形,
且$\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_NZ.$
对子流形横截相交的情形使用横截原像定理, 即有
$$
\operatorname{codim}_M (M\cap Z)=\operatorname{codim}_N Z,\quad \operatorname{codim}_N (M\cap Z)=\operatorname{codim}_N M+\operatorname{codim}_N Z.
$$
对映射横截相交的情形使用横截原像定理, 可以得到纤维积
$$
M \times_N Z:=\{(a,b)\in M\times Z|f(a)=g(b)\}
$$
为光滑流形.
证明考虑图像$\Gamma_f,\Gamma_g$与对角流形$\Delta\subset M\times N\times Z\times N,$
通过说明$(\Gamma_f\times \Gamma_g)\pitchfork \Delta,$ 投影得到结论.
稳定性
若光滑映射$f$的某种性质在任意小扰动光滑同伦$f_t$下保持($t<\varepsilon$),
那么称该性质在扰动下是稳定的.
定理 5. 紧流形$M$上的浸入/淹没$f:M\rightarrow N$是稳定的.
若$f$为浸入, 那么$\,\forall\,p\in M,$
$df_p$中存在$m\times m$阶子式行列式非零. 由连续性,
存在邻域$U_p$以及$\varepsilon_p>0,$ 使得$U_p\times [0,\varepsilon_p]$上,
$d(f_t)_q$中同样位置的$m\times m$阶子式行列式仍非零.
取$\bigcup_{i=1}^k U_i$覆盖$M,$ 记$\varepsilon=\min_i\{\varepsilon_i\},$
那么$f_t$就在$M$上是浸入, $\,\forall\,t\le \varepsilon.$
从而浸入都是稳定的, 同理淹没也是.
定理 6. $n$维光滑流形$M$的子集$S$是$M$中余维数为$l$的嵌入子流形, 当且仅当$\,\forall\,p\in S,$ $\,\exists\,$淹没$g:U\rightarrow \mathbb{R}^l,$ 使得$p\in U,$ 且$S\cap U=g^{-1}(0).$
$\Rightarrow:$ 只需取子流形坐标卡,
投影到$\mathbb{R}^{\operatorname{codim}S}$即可.
$\Leftarrow:$ 由正则值原像定理立即得到.
推论 7. 设$M$为紧致流形, 若光滑映射$f:M\rightarrow N$与$N$中闭嵌入子流形$Z$横截, 那么横截性在$f$的扰动下稳定.
记$F(x,t)=F_t.$ 由于$F^{-1}(N\setminus Z)$为开集,
$(M\setminus f^{-1}(Z))\times \{0\}\subset F^{-1}(N\setminus Z),$
$\,\forall\,p\in M\setminus f^{-1}(Z),$ $\,\exists\,U_p\subset M,$
$I_p=[0,\varepsilon_p),$ 使得$F(U_p\times I_p)\cap Z=\varnothing.$
于是$\,\forall\,t<\varepsilon_p,$ 在$U_p$上$f_t\pitchfork Z.$
由上一定理, $\,\forall\,q\in Z,$
存在淹没$g:V\rightarrow \mathbb{R}^{\operatorname{codim}Z},$
使得$Z\cap V=g^{-1}(0).$ 断言$\,\forall\,p\in f^{-1}(q),$
$g\circ f$也是淹没. 只需说明
$$
dg_q\circ df_p(T_pM)=\mathbb{R}^{\operatorname{codim}Z}.
$$
而这是因为横截性与淹没性质,
$$
df_p(T_pM)+T_qZ=T_qN,\quad dg_p(T_qN)=\mathbb{R}^{\operatorname{codim}Z},\quad dg_q(T_qZ)=0.
$$
最后的等式是因为$g(Z\cap V)=0.$
进而, $g\circ f$在$p$点邻域$\overline U_p\subset f^{-1}(V)$上为淹没.
由淹没的稳定性, 存在充分小的$\varepsilon_p>0,$
使得$g\circ f_t$在$U_p$上仍是淹没, $\,\forall\,t<\varepsilon_p$.
欲证$U_p$上$f_t\pitchfork Z,$ 要求横截性
$$
d(f_t)_p(T_pM)+T_{f(p)}Z=T_{f(p)}N,\quad \,\forall\,p\in f^{-1}_t(Z)\cap U_p.
$$
而我们有
$$
dg\circ d(f_t)_x(T_xM)=\mathbb{R}^{\operatorname{codim}Z},\quad \ker dg_q=T_qZ,\quad \,\forall\,x\in U_p,\quad q\in Z\cap V.
$$
这就给出了横截性.
由$M$的紧性, 可以取到开覆盖$M=\bigcup_{i=1}^k U_i,$
$\varepsilon=\min_i\{\varepsilon_i\}.$ 于是$\,\forall\,t<\varepsilon,$
$f_t\pitchfork Z.$ 这就给出了横截的稳定性.
我们有拓扑学中经典引理:
引理 8. 设$f:M\rightarrow N$为单的光滑浸入. 若$M$紧致, 则$f$为光滑嵌入.
利用紧性与Hausdorff性质即可. 接下来有如下稳定性定理:
定理 9. 设$M$为紧致流形, $f:M^n\rightarrow N^n$为光滑嵌入, 则嵌入性在$f$的扰动下稳定.
已知$f$在小扰动下仍为浸入. 由引理, 只需证明$\,\exists\,\varepsilon>0,$
$\,\forall\,t<\varepsilon,$ $f_t$为单射.
不然, 存在$t_i\rightarrow 0,$ $p_i\neq q_i\in M,$
$f_{t_i}(p_i)=f_{t_i}(q_i).$ 由紧性, 存在$p_i\rightarrow p_0,$
$q_i\rightarrow q_0,$ $f(p_0)=f(q_0).$ 由$f$为嵌入, $p_0=q_0=:p.$
记$G(x,t)=(f_t(x),t),$ 那么
$$
d_{(p,0)}G=\begin{bmatrix}
d_pf&*\\
0&1
\end{bmatrix}
$$
列满秩, 即$G$在$p$点为浸入.
故$G$在$p$点小邻域处为单射. 然而
$$
G(p_i,t_i)=(f_t(p_i),t_i)=(f_t(q_i),t_i)=G(q_i,t_i),\quad p_i,q_i\rightarrow p,\quad t_i\rightarrow 0.
$$
这就与小邻域矛盾.
定理 10. 设$M$为紧致流形, $f:M\rightarrow N$为微分同胚, 则微分同胚性在$f$的扰动下稳定.
微分同胚为嵌入+满射, 故只需证明满射性在扰动下稳定. 不妨设$N$为连通的,
不然依连通分支讨论即可.
由于淹没在扰动下稳定, $\,\exists\,\varepsilon>0,$
$\,\forall\,t<\varepsilon,$ $f_t$为淹没. 由淹没的典范表示,
$f_t$为开映射. 另一方面, 由$f_t$从紧集打到Hausdorff空间,
$f_t$也是闭映射. 故$f_t(M)$既开又闭, 为全空间$N.$ 这就证明了结论.
文章最后更新于 2023-06-11 14:37:16
Immersions, Submersions, and Embeddings
Immersion and submersion have canonical representations. Local constant-rank maps also have similar canonical representations.
The first two proof methods are standardization $D\widehat f,$ and then completing the dimension to construct local diffeomorphisms.
Just use this diffeomorphism to transform the coordinate system.
Immersion + homeomorphism = embedding, that is, the subspace topology of the embedded submanifold is consistent with the induced topology.
By the local representation of immersion (with the homeomorphism property), the submanifold has submanifold coordinate cards, which are slices.
regular value
For smooth mapping $f:M\rightarrow N,$, if it is submerged in $p\in M$, then $p$ is regular point,
Otherwise critical point.The image of the critical point is called critical value.
A point on $N$ that is not a critical value is called regular value.that is to say,
The original image of the regular value must be a regular point (submerged), and an empty set is allowed.
Theorem 1 (Canonical value original image theorem). Let $f:M\rightarrow N$ be a smooth map, $q\in N$ be the regular value of $f$. Then $f^{-1}(q)$ is the $(\dim M-\dim N)$-dimensional embedding submanifold.
$\,\forall\,p\in f^{-1}(q),$ is represented by submerged parts,
There is a coordinate neighborhood $(U,\varphi)\ni p,$ $(V,\psi)\ni q,$ such that
$\widehat f:\varphi(U)\rightarrow \psi(V)$ is projection mapping, $\varphi(p)=0,$
$\psi(q)=0$. So
$$
\widehat f^{-1}(0)=\{(0,y)|y\in \mathbb{R}^{\dim M-\dim N}\}=\varphi\circ f^{-1}(q).
$$
This gives $f^{-1}(q)$ as the submanifold coordinate card $(U\cap f^{-1}(q),{\color{red}\pi_y}\circ \varphi)$ of the $(\dim M-\dim N)$ dimensional slice.
Theorem 2 (Disc Lemma). Let $M,N$ be a smooth manifold with the same dimensions, and $M$ be compact. Let $f:M\rightarrow N$ be a smooth map. If $q\in N$ is the regular value of $f$, then $f^{-1}(q)\subset M$ is a finite point set $\{p_i\}_{i=1}^k$, and there is an open neighborhood $V,$ of $q$ in $N$ such that $f^{-1}(V)=\bigsqcup_{i=1}^k U_i,$ $p_i\in U_i,$ and $f|_{U_i}$ are diffeomorphisms.
According to the canonical value original image theorem, $f^{-1}(q)=\{p_i\}$ is a $0$-dimensional submanifold,
There is a submanifold coordinate neighborhood $W_i,$ that has no intersection. Since $M$ is compact,
If $\{p_i\}$ is an infinite point set, there will be a gathering point, but such a coordinate neighborhood $W_i$ cannot be obtained at the gathering point.
Or $f^{-1}(q)$ is a closed set, so it is compact.
Since $f$ is both immersed and submerged, it is a local diffeomorphism.
Let’s assume it is diffeomorphism on each $W_i$. Next, take
$$
V=\bigcap_{i=1}^k f(W_i) \setminus f(M\setminus \bigcup_{i=1}^k W_i).
$$
Where $M\setminus \bigcup_{i=1}^k W_i$ is a closed set, and a closed set in a compact set is compact.
Therefore $f(M\setminus \bigcup_{i=1}^k W_i)$ is a compact set, and thus a closed set.
So $V$ is indeed an open neighborhood, and $f^{-1}(V)\subset \bigcup_{i=1}^k W_i.$
Since $f|_{W_i}$ is diffeomorphism,
$\,\exists\,U_i{\color{red}{(=f^{-1}(V)\cap W_i)} }\subset W_i,$
Let $f^{-1}(V)=\bigsqcup_{i=1}^k U_i,$ and $f|_{U_i}$ be diffeomorphisms.
Theorem 3 (Sard’s theorem). The critical value set of the smooth map $f:M\rightarrow N$ is the zero measurement set, which is almost everywhere a regular value (thus dense).
cross section
Let $f:M\rightarrow N$ be a smooth map and $Z\subset N$ be an embedded submanifold.
If $\,\forall\,p\in f^{-1}(Z),$
$$
df_p(T_pM)+T_{f(p)}Z=T_{f(p)}N,
$$
Then $f$ and $Z$ are said to be transverse, denoted as $f\pitchfork Z.$. When $Z$ is a single point set,
$Z$ is the regular value of $f$.
In particular, if $M\subset N$ is an embedded submanifold, then it is called $M\pitchfork Z$,
If you put the mappings $\iota:M\rightarrow N$ and $Z$ transversally. At this time, there is
$$
T_pM+T_pZ=T_pN,\quad \,\forall\,p\in M\cap Z.
$$
Similarly, two mapping transversals can also be defined, in which case the mapping is not necessarily required to be embedded.
Theorem 4 (Transversal original image theorem). Let $f:M\rightarrow N$ be a smooth map, $Z\subset N$ be an embedded submanifold. If $f\pitchfork Z,$, then $f^{-1}(Z)$ is the embedded submanifold in $M$, and $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_N Z.$
Since $Z$ is an embedded submanifold, $\,\forall\,q\in Z,$ has coordinate cards $(V,\psi),$ $q\in V,$
Let $\psi(Z\cap V)\subset \mathbb{R}^{\dim Z}$ be a slice.
Get projection$\pi:\mathbb{R}^{n}\rightarrow \mathbb{R}^{\operatorname{codim}_N Z},$
Then $\psi (Z\cap V)=\pi^{-1}(0).$
Assert that $g:=\pi \circ \psi \circ f:f^{-1}(V)\rightarrow \mathbb{R}^{\operatorname{codim}_NZ}$ takes $0$ as a regular value.
Just say $\,\forall\,p\in f^{-1}(Z)\cap f^{-1}(V)=f^{-1}(Z\cap V),$
$dg_p$ is submerged. According to the transversality condition,
$$
d\psi \circ df_p(T_pM)+ T_{\psi\circ f(p)}\psi(Z)=T_{\psi\circ f(p)}\psi(N)=\mathbb{R}^n,\quad T_{\psi\circ f(p)}\psi(Z)=\mathbb{R}^{\dim Z}.
$$
Apply $d\pi=\pi,$ again to have
$$
dg_p(T_pM)=d\pi\circ d\psi\circ df_p(T_pM)=\mathbb{R}^{\operatorname{codim}Z}.
$$
According to the canonical value original image theorem,
${\color{red}{g^{-1}(0)=f^{-1}(Z\cap V)=f^{-1}(Z)\cap f^{-1}(V)} }$ is the embedded submanifold,
and
$$
\dim g^{-1}(0)=\dim M-\operatorname{codim}_NZ,\quad \operatorname{codim}_M (f^{-1}(Z)\cap f^{-1}(V))=\operatorname{codim}_NZ.
$$
Due to the arbitrariness of $q\in V$, $f^{-1}(Z)$ is an embedded submanifold,
And $\operatorname{codim}_M f^{-1}(Z)=\operatorname{codim}_NZ.$
For the case where submanifolds intersect transversely, we use the transversal preimage theorem, that is, we have
$$
\operatorname{codim}_M (M\cap Z)=\operatorname{codim}_N Z,\quad \operatorname{codim}_N (M\cap Z)=\operatorname{codim}_N M+\operatorname{codim}_N Z.
$$
Using the transversal preimage theorem for the case where mapping transversals intersect, the fiber product can be obtained
$$
M \times_N Z:=\{(a,b)\in M\times Z|f(a)=g(b)\}
$$
is a smooth manifold.
Proof Consider the image $\Gamma_f,\Gamma_g$ and the diagonal manifold $\Delta\subset M\times N\times Z\times N,$
The conclusion is reached by stating the $(\Gamma_f\times \Gamma_g)\pitchfork \Delta,$ projection.
Stability
If a certain property of the smooth map $f$ is maintained under any small perturbation of the smooth homotopy $f_t$ ($t<\varepsilon$),
Then this property is said to be under perturbation stable.
Theorem 5. Immersion/submergence $f:M\rightarrow N$ on a compact manifold $M$ is stable.
If $f$ is immersion, then $\,\forall\,p\in M,$
There is a non-zero determinant of the subformula of order $m\times m$ in $df_p$. By continuity,
There are neighbors $U_p$ and $\varepsilon_p>0,$ such that on $U_p\times [0,\varepsilon_p]$,
The determinant of the subformula of $m\times m$ at the same position in $d(f_t)_q$ is still non-zero.
Take $\bigcup_{i=1}^k U_i$ to cover $M,$ and write $\varepsilon=\min_i\{\varepsilon_i\},$
Then $f_t$ is immersed on $M$, $\,\forall\,t\le \varepsilon.$
Therefore, immersion is stable, and so is submersion.
Theorem 6. The subset $S$ of the $n$-dimensional smooth manifold $M$ is an embedded submanifold with codimension $l$ in $M$, if and only if $\,\forall\,p\in S,$ $\,\exists\,$ submerges $g:U\rightarrow \mathbb{R}^l,$ such that $p\in U,$ and $S\cap U=g^{-1}(0).$
$\Rightarrow:$ Just take the submanifold coordinate card,
Just project it to $\mathbb{R}^{\operatorname{codim}S}$.
$\Leftarrow:$ is obtained immediately from the canonical value original image theorem.
Corollary 7. Assume $M$ is a compact manifold. If the closed embedded submanifold $Z$ in the smooth mapping $f:M\rightarrow N$ and $N$ is transverse, then the transversality is stable under the perturbation of $f$.
Note $F(x,t)=F_t.$ Since $F^{-1}(N\setminus Z)$ is an open set,
$(M\setminus f^{-1}(Z))\times \{0\}\subset F^{-1}(N\setminus Z),$
$\,\forall\,p\in M\setminus f^{-1}(Z),$ $\,\exists\,U_p\subset M,$
$I_p=[0,\varepsilon_p),$ makes $F(U_p\times I_p)\cap Z=\varnothing.$
So $\,\forall\,t<\varepsilon_p,$ is on $U_p$ and $f_t\pitchfork Z.$
From the above theorem, $\,\forall\,q\in Z,$
Existential drowning $g:V\rightarrow \mathbb{R}^{\operatorname{codim}Z},$
make$Z\cap V=g^{-1}(0).$ assert$\,\forall\,p\in f^{-1}(q),$
$g\circ f$ is also submerged. Just to clarify
$$
dg_q\circ df_p(T_pM)=\mathbb{R}^{\operatorname{codim}Z}.
$$
And this is because of the transversal and submerged nature,
$$
df_p(T_pM)+T_qZ=T_qN,\quad dg_p(T_qN)=\mathbb{R}^{\operatorname{codim}Z},\quad dg_q(T_qZ)=0.
$$
The final equation is because $g(Z\cap V)=0.$
Furthermore, $g\circ f$ is submerged in the point neighborhood $\overline U_p\subset f^{-1}(V)$ of $p$.
By submerged stability, there is a sufficiently small $\varepsilon_p>0,$
So that $g\circ f_t$ is still submerged on $U_p$, $\,\forall\,t<\varepsilon_p$.
To prove $U_p$ on $f_t\pitchfork Z,$ requires transversality
$$
d(f_t)_p(T_pM)+T_{f(p)}Z=T_{f(p)}N,\quad \,\forall\,p\in f^{-1}_t(Z)\cap U_p.
$$
And we have
$$
dg\circ d(f_t)_x(T_xM)=\mathbb{R}^{\operatorname{codim}Z},\quad \ker dg_q=T_qZ,\quad \,\forall\,x\in U_p,\quad q\in Z\cap V.
$$
This gives transversality.
From the compactness of $M$, we can get the open covering $M=\bigcup_{i=1}^k U_i,$
$\varepsilon=\min_i\{\varepsilon_i\}.$ So $\,\forall\,t<\varepsilon,$
$f_t\pitchfork Z.$ This gives cross-sectional stability.
We have the classic lemma in topology:
Lemma 8. Let $f:M\rightarrow N$ be a single smooth immersion. If $M$ is compact, then $f$ is a smooth embedment.
Just use compactness and Hausdorff property. Next there is the following stability theorem:
Theorem 9. Assume $M$ is a compact manifold and $f:M^n\rightarrow N^n$ is a smooth embedding, then the embeddedness is stable under the perturbation of $f$.
It is known that $f$ is still immersed under small perturbations. According to the lemma, we only need to prove $\,\exists\,\varepsilon>0,$
$\,\forall\,t<\varepsilon,$ $f_t$ is a single shot.
Otherwise, there exists $t_i\rightarrow 0,$ $p_i\neq q_i\in M,$
$f_{t_i}(p_i)=f_{t_i}(q_i).$ Due to compactness, there exists $p_i\rightarrow p_0,$
$q_i\rightarrow q_0,$ $f(p_0)=f(q_0).$ is embedded from $f$, $p_0=q_0=:p.$
Remember $G(x,t)=(f_t(x),t),$ Then
$$
d_{(p,0)}G=\begin{bmatrix}
d_pf&*\\
0&1
\end{bmatrix}
$$
The column is full rank, that is, $G$ is immersed at point $p$.
Therefore, $G$ is injective in the small neighborhood of $p$. However,
$$
G(p_i,t_i)=(f_t(p_i),t_i)=(f_t(q_i),t_i)=G(q_i,t_i),\quad p_i,q_i\rightarrow p,\quad t_i\rightarrow 0.
$$
This is in conflict with small neighborhoods.
Theorem 10. Assume $M$ is a compact manifold and $f:M\rightarrow N$ is a diffeomorphism, then the diffeomorphism is stable under the perturbation of $f$.
Diffeomorphism is embedding + surjection, so we only need to prove that surjection is stable under perturbation. Let $N$ be connected,
Otherwise, just follow the connected component discussion.
Since submersion is stable under disturbance, $\,\exists\,\varepsilon>0,$
$\,\forall\,t<\varepsilon,$ $f_t$ is submerged. It is represented by the model of submergence,
$f_t$ is an open mapping. On the other hand, $f_t$ is used to map from a compact set to a Hausdorff space,
$f_t$ is also a closed map. Therefore, $f_t(M)$ is both open and closed, and is the whole space $N.$. This proves the conclusion.
The article was last updated on 2023-06-11 14:37:16
