User:Inconsistency/Sandbox

Proof 1
Let $f \colon S \hookrightarrow T$, $g \colon T \hookrightarrow S$ as in the assumptions.

Using these injections one attempts to construct partitions:
 * $\mathfrak S \subseteq \mathcal P\left(S\right)$ and $\mathfrak T \subseteq \mathcal P\left(T\right)$

as well as a bijection:
 * $F \colon \mathfrak S \hookrightarrow \hspace{-8pt} \rightarrow \mathfrak T$

and a family of bijections:
 * $\forall \mathfrak s\in\mathfrak S \colon \left(f_{\mathfrak s} \colon \mathfrak \hookrightarrow \hspace{-8pt} \rightarrow F\left(\mathfrak s\right)\right)$

Now there exists a bijection $\mathbf{B \colon S \hookrightarrow \hspace{-6pt} \rightarrow T}$


 * Define $\displaystyle B := \bigcup_{\mathfrak s \mathop\in \mathfrak S} f_{\mathfrak s}$


 * Since $F$ is an injection: $\forall \mathfrak s_1, \mathfrak s_2 \in \mathfrak S \colon \mathfrak s_1 \ne \mathfrak s_2 \implies F(\mathfrak s_1) \ne F\left(\mathfrak s_2\right) \implies F(\mathfrak s_1) \cap F(\mathfrak s_2) = \varnothing$
 * Since $F$ is a surjection: $\forall \mathfrak t \in \mathfrak T \colon \exists \mathfrak s \in \mathfrak S \colon F(\mathfrak s) = \mathfrak t$
 * Now by
 * Union of Left-Total Relations is Left-Total
 * Union of Many-to-One Relations with Disjoint Domains is Many-to-One
 * Union of One-to-Many Relations with Disjoint Images is One-to-Many
 * $\left\{\mathfrak s\right\}_{\mathfrak s \mathop\in \mathfrak S}$ are mutually disjoint $\implies B$ is functional
 * $S = \bigcup \mathfrak S \implies B$ is left-total
 * $\left\{\mathfrak t\right\}_{\mathfrak t \mathop\in \mathfrak T}$ are mutually disjoint and $F$ is injective $\implies B$ is injective
 * $T = \bigcup \mathfrak T$ and $F$ is surjective $\implies B$ is surjective

Definition of the Partitions
By Composite of Injections is Injection: $(h_S := g\circ f) : S \hookrightarrow S$ and $(h_T := f\circ g) : T \hookrightarrow T$. In the following denote by $X$ either $S$ or $T$ and define $\preceq_X \subseteq X^2$ by
 * $\langle x_1,x_2 \rangle \in \preceq_X \hspace{8pt}\Leftrightarrow\hspace{8pt} \exists n\in\N: x_2 = h_X^n x_1$

$\mathbf{\preceq_X}$ is a preorder relation


 * (Reflexivity) Let $x \in X$ then $x = h_X^0 x$ $\implies x\preceq x$
 * (Transitivity) Let $x \preceq y$ and $y \preceq z$ then $y = h_X^m x$ and $z = h_X^n y$ $\implies z = h_X^{m+n}x \implies x\preceq z$

Now define $\sim_X \subseteq X^2$ by
 * $\langle x_1,x_2 \rangle \in \sim_X \hspace{8pt}\Leftrightarrow\hspace{8pt} x_1\preceq x_2 \vee x_2\preceq x_1$

$\mathbf{\sim}$ is an equivalence relation


 * (Reflexivity) and (Symmetry) are trivial consequences of the definition of $\sim_X$ and the reflexivity of $\preceq_X$.
 * (Transitivity) Let $x_1,x_2,x_3\in X$ such that $x_1,x_2\preceq x_3$. Then $x_3 = h_X^m x_1 = h_X^n x_2$ for $n,m \in \N$.
 * Wlog assume $m\le n$. Then $x_1 = h_X^{n-m} x_2 \underset{h_X\text{ injective}}\iff h_X^m x_1 = h_X^m h_X^{n-m} x_2 \iff x_3 = h_X^n x_2$.
 * Therefore $x_1 \preceq x_2$.

Define $\mathfrak X := X/\!\!\sim$. Now the Fundamental Theorem on Equivalence Relations implies that the equivalence classes $\mathfrak x \in \mathfrak X$ form a partition of $X$.

Definition of $F$
Denote by $\left[\!\left[{s}\right]\!\right]$ the equivalence class of $s\in S$ and define $F$ by $F\left(\left[\!\left[{s}\right]\!\right]\right) := \left[\!\left[{f\left(s\right)}\right]\!\right]$

$\mathbf F$ is well-defined and injective
 * Since Equivalence Class is not Empty there exists $s\in \mathfrak s$ and $\left[\!\left[{f\left(s\right)}\right]\!\right]$ exists as well.
 * Now it remains to show the independence of the chosen element.
 * $s_1,s_2\in \left[\!\left[{s}\right]\!\right]_S \iff s_1\sim_S s_2 \underset{\text{wlog}}\iff s_1 \preceq s_2 \overset{n\in\N}\iff s_2 = h_S^n s_1 = (g\circ f)^n s_1 \underset{f\text{ injective}}\iff f(s_2) = f((g\circ f)^n s_1) = f\circ(g\circ f)^n s_1 = (f\circ g)^n\circ f\ s_1 = h_T^n (f(s_1)) \iff f(s_1) \preceq_T f(s_2) \iff f(s_1) \sim_T f(s_2)$

$\mathbf F$ is surjective
 * Let $\left[\!\left[{t}\right]\!\right]_T \in T/\!\sim_T$ then $\widetilde{g}\left[\!\left[{t}\right]\!\right]_S \in S/\!\sim_S$ and obviously $\widetilde{f}\widetilde{g}(\left[\!\left[{t}\right]\!\right]_S) = \left[\!\left[{f(g(t))}\right]\!\right]_T = \left[\!\left[{h_T(t)}\right]\!\right]_T = \left[\!\left[{t}\right]\!\right]_T$.

Definition of the $f_{\mathfrak s}$
Let $\mathfrak s\in \mathfrak S$ and denote by $h_{\mathfrak x} := h_X\!\!\restriction_{\mathfrak x}$.

$\mathbf{\mathfrak s}$ is totally ordered
 * $r,s\in\mathfrak s \iff r \sim s \iff r\preceq s \vee s\preceq r$

s is a chain

Since Equivalence Class is not Empty and since s is a chain there exist $r \preceq s\in\mathfrak s$ that is $\exists n\in\N \colon s = h_{\mathfrak s}^n r$. Now consider the family of functions $p_s \colon \mathfrak s \to \Z$ defined by
 * $p_s\left(r\right) :=

\begin{cases} n & s = h_{\mathfrak s}^n r \\ -n & r = h_{\mathfrak s}^n s \end{cases} $

$\mathbf{s = h_{\mathfrak s}^n s \implies h_{\mathfrak s}^n = \operatorname{id}}$
 * Let $r\in\mathfrak s$ then by totality $r \preceq s \vee s\preceq r$.
 * Assume $r\preceq s$
 * Then $s = h_{\mathfrak s}^k r$ for some $k\in\N$.
 * And $r = h_{\mathfrak s}^n r \underset{h_{\mathfrak s}\text{ injective}}\iff h_{\mathfrak s}^k r = h_{\mathfrak s}^{n+k} r \iff s = h_{\mathfrak s}^n s$
 * Assume $s\preceq r$
 * Then $r = h_{\mathfrak s}^k s$ for some $k\in\N$.
 * And $r = h_{\mathfrak s}^n r \iff h_{\mathfrak s}^k s = h_{\mathfrak s}^{n+k} s \iff h_{\mathfrak s}^k s = h_{\mathfrak s}^k s$
 * So $h_{\mathfrak s}^n r = r$ for all $r \in\mathfrak s$. Therefore $h_{\mathfrak s}^n = \operatorname{id}$

Therefore either $\exists n\in\N \colon h_{\mathfrak s}^n = \operatorname{id}$ or $\forall $

$\mathbf{\langle\mathfrak s, \preceq\rangle \simeq \langle\Z/n\Z, \le\rangle}$ for some $\mathbf{n\in\N}$


 * Now one can differentiate the following two cases:
 * $\exists n\in\N \colon h_{\mathfrak s}^n = \operatorname{id}$

$\mathbf{h_{\mathfrak s}^n = \operatorname{id} \iff h_{F\left(\mathfrak s\right)}^n = \operatorname{id}}$

Therefore there exists $f_{\mathfrak s} \colon \mathbf{\langle \mathfrak s, \preceq\rangle \simeq \langle\Z/n\Z, \le\rangle \simeq \langle F\left(\mathfrak s\right), \preceq\rangle}$

$\mathbf{\mathfrak s}$ is a chain
 * (Totality) Let $r,s\in\mathfrak s$ then $r\sim s \implies r \preceq s \vee s\preceq r$.
 * (Antisymmetry) Let $r\ne s\in\mathfrak s$ then

s \in \left(h_S^n\right)^{-1}(r) for some r \implies f(s) \in h_T^n^-1(f(r)).

Since Equivalence Class is not Empty one can choose $r,s\in\mathfrak s$ and differentiate between the following two cases:


 * $\exists r,s\in\mathfrak s; n\in\N \colon s = h_S^n r$
 * Now $r,y\in\left[\!\left[x\right]\!\right]_X \implies r\sim y \implies r \preceq y \vee y \preceq r$.
 * By an abuse of notation write $y = h_X^k r$ for $k\in\Z$ and keep in mind that for $k<0$ $y = h_X^k r$ means $r = h_X^{-k} y$.
 * Consider two cases:


 * Define $n := l-k$ the smallest of such distances (which exists becaus $\N^2$ is well ordered). Then for any $y\in\left[\!\left[x\right]\!\right]_X : y = h_X^n y$.
 * Now define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z/n\Z$ by $o_r(y) = k + n\Z$.
 * $\not\exists l\ne k: y=h_X^l r$
 * Define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z = \Z/0\Z$ by $o_r(y) = k + 0\Z$.
 * Observe that $o_r$ is injective and surjective by construction and trivially order preserving.


 * Otherwise

Definition of the Partitions

 * $\widetilde X_0 := X$
 * $\widetilde X_{n+1} := h_X \widetilde X_n$


 * $\displaystyle X_0 := \bigcap_{n\mathop\in\N} X_n$
 * $X_{n+1} := \widetilde X_{n+1} \setminus X_n$

Define $\mathcal X := \left\{X_n\right\}_{n\mathop\in\N}$

$\mathbf{\bigsqcup \mathcal X = X}$
 * Let $m -n

r, h_X == N ohne repetitionen r, h_X == Z/nZ mit -*- => Z/nZ <= [x] Aber genauso [x] <= Z Theorem: Alle Subtypen von Z haben diese Form. Lemma: f ist ein morphismus von typen. (trivial) Theorem: Seien S,T typen und S<=T, T<=S. Dann S==T.

Betrachte die totalordnung auf [x]. 1) [x] besitzt ein minimales element => [x]

General construction: Let v : Z -> X for some set Z be injective st h|_{Im(v)} bijective.

wie wählt man Z intelligent, sodass entweder NS == S oder N = 0 ???

Proof

 * By Equivalence Class is not Empty one can fix an element $r\in[x]$.
 * And define addition ...
 * This definition does not depend on the choice of $r$ since if $q$ is another representative:
 * $q = h_X^k r$ => (x +_q y) = ... = (x +_r y)
 * Note that this addition is commutative by construction.
 * Finally observe that $\forall x,y,a:x ...

Now [f]:[s]->[t] is a monomorphism of groups and is completely determined by [f](r). [g] analogously. But [f] monomorph => ord([s])<=ord([t]) and ... => ord([s])=ord([t]) => [s] \simeq [t]

Since Equivalence Classes are Disjoint these $\left[\!\left[{x}\right]\!\right]_X$ form a partition of $X$ and therefor every element of $X$ belongs to exactly one element of $X/\!\sim_X$ (namely $\left[\!\left[{x}\right]\!\right]_X$).

Now define $\widetilde{f} : S/\!\sim_S \to T/\!\sim_T$ by $f(\left[\!\left[{s}\right]\!\right]_S) = \left[\!\left[{f(s)}\right]\!\right]_T$ and $\widetilde{g}$ analogous.

Claim

 * $\langle\left[\!\left[{x}\right]\!\right]_X, \preceq\rangle \simeq \langle\Z/n\Z, \le\rangle$ for some $n\in\N$

Proof

 * By Equivalence Class is not Empty one can fix $r\in\left[\!\left[x\right]\!\right]_X$.
 * Let $y \in \left[\!\left[x\right]\!\right]_X$ be arbitrary.
 * Now $r,y\in\left[\!\left[x\right]\!\right]_X \implies r\sim y \implies r \preceq y \vee y \preceq r$.
 * By an abuse of notation write $y = h_X^k r$ for $k\in\Z$ and keep in mind that for $k<0$ $y = h_X^k r$ means $r = h_X^{-k} y$.
 * Consider two cases:
 * $\exists l\ne k: y=h_X^l r$
 * Wlog assume $k<l$ then $h_X^{l-k} r = r \underset{h_X\text{ injective}}\iff h_X^kh_X^{l-k} r = h_X^kr \iff h_X^kr = h_X^kr$.
 * Define $n := l-k$ the smallest of such distances (which exists becaus $\N^2$ is well ordered). Then for any $y\in\left[\!\left[x\right]\!\right]_X : y = h_X^n y$.
 * Now define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z/n\Z$ by $o_r(y) = k + n\Z$.
 * $\not\exists l\ne k: y=h_X^l r$
 * Define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z = \Z/0\Z$ by $o_r(y) = k + 0\Z$.
 * Observe that $o_r$ is injective and surjective by construction and trivially order preserving.

= Notation:Differential_Geometry =

Propose this new segment.

Let in the following $M$ be a smooth manifold.

$\Omega^n(M)$ - the set of all smooth n-forms on M