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1. • CommentRowNumber1.
   • CommentAuthorDexter Chua
   • CommentTimeAug 6th 2016
   • (edited Aug 6th 2016)
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   Author: Dexter Chua
   Format: MarkdownItexRestructured the [[manifold]] entry to avoid duplication with [[pseudogroup]], and moved the section on the tangent bundle to [[tangent bundle]]

   Restructured the manifold entry to avoid duplication with pseudogroup, and moved the section on the tangent bundle to tangent bundle

2. • CommentRowNumber2.
   • CommentAuthorJoshua Meyers
   • CommentTimeAug 31st 2019
   • (edited Aug 31st 2019)
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   Author: Joshua Meyers
   Format: MarkdownItexThe definition via pseudogroups is supposed to have the following problem: > There is in general no notion of morphisms between manifolds. At best, we can only talk about isomorphisms of manifolds. However, couldn't this problem be solved by using a "pseudomonoid" instead of a pseudogroup? We can define a pseudomonoid $M$ on the topological space $X$ as a wide subcategory of $\text{Op}(X)$ such that for all $U, V, U_\alpha$ open in $X$, $f: U\rightarrow V$ such that $(U_\alpha)$ covers $U$, we have the condition $f\in M \iff (\forall \alpha) f|_{U_\alpha}:U\cap U_\alpha\rightarrow V\cap f(U_\alpha) \in M$.

   The definition via pseudogroups is supposed to have the following problem:

   There is in general no notion of morphisms between manifolds. At best, we can only talk about isomorphisms of manifolds.

   However, couldn’t this problem be solved by using a “pseudomonoid” instead of a pseudogroup? We can define a pseudomonoid $M$ on the topological space $X$ as a wide subcategory of $\text{Op}(X)$ such that for all $U, V, U_\alpha$ open in $X$, $f: U\rightarrow V$ such that $(U_\alpha)$ covers $U$, we have the condition $f\in M \iff (\forall \alpha) f|_{U_\alpha}:U\cap U_\alpha\rightarrow V\cap f(U_\alpha) \in M$.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 1st 2019
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   Author: Todd_Trimble
   Format: MarkdownItexI'm not sure right away. I'd have to think about it.

   I’m not sure right away. I’d have to think about it.

4. • CommentRowNumber4.
   • CommentAuthorJoshua Meyers
   • CommentTimeSep 2nd 2019
   • (edited Sep 2nd 2019)
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   Author: Joshua Meyers
   Format: MarkdownItexHere's a new way of characterizing manifolds that I like better. Let $X$ be a topological space and let $N$ be the monoid of partial continuous functions $X\rightarrow X$ (any element of $N$ has open domain by continuity), ordered by inclusion of graphs. Then define a pseudomonoid on $X$ to be a submonoid $S$ of $N$ such that * If $f, f_\alpha\in N$ and $f=\cup_\alpha f_\alpha$, then $f\in S$ iff $(\forall \alpha) (f_\alpha\in S)$. (Equivalently, $S$ is a lower set of $N$ and the inclusion $S\rightarrow N$ lifts joins.) (This condition implies that $S$ is closed under open restriction, and, being a submonoid of $N$, $S$ contains the identity map $\text{id}_X$, so $S$ contains all coreflexive elements of $N$.) Let $G$ be the inverse semigroup $G$ of all elements of $S$ that are bijective and whose opposite (as a relation) is also in $S$. Now let $M$ be a set. A chart on $M$ is a partial bijection $M\rightarrow X$. Two charts $a,b$ are $S$-compatible iff $b^{\text{op}}a\in G$ (or equivalently, $b^{\text{op}}a, a^{\text{op}}b \in S$). (A chart is compatible with itself iff its image is open.) An $S$-atlas on $M$ is a collection of pairwise $S$-compatible maps whose domains cover $M$. An $S$-manifold is a set $M$ equipped with an $S$-atlas. A map $f$ between $S$-manifolds $M$, $N$ with $S$-atlases $A$, $B$ is a morphism iff there exists a collection of pairs $(\phi_\alpha, \psi_\alpha)\in A\times B$ such that the collection $\{\text{dom} \phi_\alpha \times \text\{dom} \psi_\alpha\}$ covers the graph of $f$ and for all $\alpha$, $\psi_\alpha f \phi_\alpha^{\text{op}}\in S$. By using partial functions we don't need to worry about having all different domains involved and keeping track of restrictions and such.

   Here’s a new way of characterizing manifolds that I like better.

   Let $X$ be a topological space and let $N$ be the monoid of partial continuous functions $X\rightarrow X$ (any element of $N$ has open domain by continuity), ordered by inclusion of graphs. Then define a pseudomonoid on $X$ to be a submonoid $S$ of $N$ such that

   • If $f, f_\alpha\in N$ and $f=\cup_\alpha f_\alpha$, then $f\in S$ iff $(\forall \alpha) (f_\alpha\in S)$. (Equivalently, $S$ is a lower set of $N$ and the inclusion $S\rightarrow N$ lifts joins.)

   (This condition implies that $S$ is closed under open restriction, and, being a submonoid of $N$, $S$ contains the identity map $\text{id}_X$, so $S$ contains all coreflexive elements of $N$.)

   Let $G$ be the inverse semigroup $G$ of all elements of $S$ that are bijective and whose opposite (as a relation) is also in $S$.

   Now let $M$ be a set. A chart on $M$ is a partial bijection $M\rightarrow X$. Two charts $a,b$ are $S$-compatible iff $b^{\text{op}}a\in G$ (or equivalently, $b^{\text{op}}a, a^{\text{op}}b \in S$). (A chart is compatible with itself iff its image is open.)

   An $S$-atlas on $M$ is a collection of pairwise $S$-compatible maps whose domains cover $M$.

   An $S$-manifold is a set $M$ equipped with an $S$-atlas. A map $f$ between $S$-manifolds $M$, $N$ with $S$-atlases $A$, $B$ is a morphism iff there exists a collection of pairs $(\phi_\alpha, \psi_\alpha)\in A\times B$ such that the collection $\{\text{dom} \phi_\alpha \times \text\{dom} \psi_\alpha\}$ covers the graph of $f$ and for all $\alpha$, $\psi_\alpha f \phi_\alpha^{\text{op}}\in S$.

   By using partial functions we don’t need to worry about having all different domains involved and keeping track of restrictions and such.

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 2nd 2019
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   Author: Todd_Trimble
   Format: MarkdownItexJoshua, is this in the literature someplace?

   Joshua, is this in the literature someplace?

6. • CommentRowNumber6.
   • CommentAuthorJoshua Meyers
   • CommentTimeSep 3rd 2019
   • (edited Sep 3rd 2019)
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   Author: Joshua Meyers
   Format: MarkdownItexNot that I know of, but I haven't looked very hard.

   Not that I know of, but I haven’t looked very hard.