Add category of compact Hausdorff spaces#160
Add category of compact Hausdorff spaces#160dschepler wants to merge 7 commits intoScriptRaccoon:mainfrom
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Hmm, for the coregular property, I have two possible approaches: |
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Thanks for the PR! For the non-existence of NNO, and hence showing that CompHaus is not countably distributive, I suggest to use the lemma If an NNO exists, it has to be needs to be a split monomorphism. But it is clearly injective and has dense image. So it would actually be an isomorphism. I don't think that this is true when |
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I think coregularity should be easy to prove directly. Don't go via C*-algebras here. |
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The question you raise regarding a consequence of NNO existing also happens to be a special case of whether the category has cartesian filtered colimits, for the special case of the colimit |
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Regarding the property of having cofiltered-limit-stable epimorphisms: It's interesting that the counterexample from Set and Haus fails in CompHaus. In fact, by the intersection theorem, any such example where it's asking about an intersection of a codirected family of nonempty compact subobjects to the constant 1 is doomed to failure. I don't have any ideas on the general case, though. |
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How about this: if Maybe constructing the counterexample of a sequence in |
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I think I've got it now: if a natural numbers object |
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Epimorphisms might actually be stable under cofiltered limits. This is equivalent to: monomorphisms of commutative unital C*-algebras ( = injective *-homomorphisms) are stable under filtered colimits – which looks correct, even for all C*-algebras. More generally, exact cofiltered limits are likely. Also, locally copresentable looks promising. But let's try to find a purely topological proof. |
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I've found a proof of cofiltered-limit-stable epis. A brief outline: First, a lemma that any cofiltered limit of nonempty compact Hausdorff spaces is nonempty. To see this, consider the product, and for For the main statement: just apply the lemma to the cofiltered limit of inverse images of I'm still not sure whether this is just a special case of a more general argument in disguise, where the more general argument establishes exact cofiltered limits. |
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I think I'm getting closer to a proof that it's The basic idea: first prove that It should then follow that all countable powers of |
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I found an interesting paper: https://arxiv.org/pdf/1808.09738 |
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https://doi.org/10.1016/j.topol.2019.02.033 proves several results about locally copresentable categories of spaces (called: dually locally presentable categories). See Theorem 3.4 and Theorem 4.9. It appears that CompHaus is never mentioned explicitly, but I assume this is because the authors are way past that example :D. I will find a better reference. Close catch: https://arxiv.org/pdf/1508.07750 - references in particular the result that CompHausop is monadic over Set. So I assume the question is if this monad is accessible. |
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Apparently, Isbell proved that CompHausop is equivalent to the category of functors J. Isbell. Generating the algebraic theory of C(X). Algebra Universalis, 15(2):153–155, 1982 I don't have access to the paper. And tbh the papers of Isbell are never easy to understand. So I would appreciate if we can find a more direct argument for local |
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The introduction of http://www.tac.mta.ca/tac/volumes/33/12/33-12.pdf gives useful references.
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To prove a category has exact cofiltered limits, is it sufficient to show that cofiltered limits commute with binary coproducts, initial objects, and coequalizers? Or is there something subtle there that I'm missing? |
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I think I have a proof that the functor Hom(-, [0,1]) : CompHaus^op -> Set is monadic, using the crude monadicity criterion: suppose we have a coreflexive equalizer As for showing it is conservative: suppose Oh, and for having a left adjoint: that's the functor Combined with |
Yes this works as soon as cofiltered limits and finite colimits exist, otherwise they cannot be computed pointwise e.g. and this is also why it was an assumption for the property in the first place. Also, initial objects are never a problem, as the constant diagram with value X on a non-empty category has limit X. |
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I would like to hear your feedback regarding the current state of #164 (the biggest PR I have worked on so far). For this I have converted the definition of CompHaus and a selection of its properties from this PR into a yaml file id: CompHaus
name: category of compact Hausdorff spaces
notation: $\CompHaus$
objects: compact Hausdorff spaces
morphisms: continuous functions
description: |
This is the full subcategory of $\Top$ consisting of those spaces that are [compact](https://en.wikipedia.org/wiki/Compact_space) and [Hausdorff](https://en.wikipedia.org/wiki/Hausdorff_space).
nlab_link: https://ncatlab.org/nlab/show/empty+category
tags:
- topology
related_categories:
- Top
- Haus
satisfied_properties:
- property_id: locally small
reason: This is trivial.
- property_id: generator
reason: The one-point space is a generator because it represents the forgetful functor to $\Set$, which is faithful.
- property_id: products
reason: |
By the Tychonoff product theorem, a product in $\Top$ of compact Hausdorff spaces is compact; it is also clearly Hausdorff.
Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.
- property_id: equalizers
reason: |
The equalizer in $\Top$ of two continuous functions $f, g : X \rightrightarrows Y$ between compact Hausdorff spaces is a closed subspace of $X$, and therefore it is also compact Hausdorff. Since the forgetful functor from $\CompHaus$ to $\Top$ is fully faithful, this limit is reflected in $\CompHaus$ as well.
- property_id: cocomplete
reason: |
$\CompHaus$ is a reflective subcategory of $\Top$, with the reflector being the Stone-Čech compactification functor.
See [nLab](https://ncatlab.org/nlab/show/compact+Hausdorff+space#StoneCechCompactification) for example.
Therefore, as usual, we can form colimits in $\CompHaus$ by forming colimits in $\Top$ and then applying Stone-Čech compactification.
unsatisfied_properties:
- property_id: natural numbers object
reason: |
Let $I := [0, 1]$. If a natural numbers object $(N, z : 1 \to N, s : N \to N)$ existed, then we could iterate the initial conditions $I\to I\times I$, $x \mapsto (x, x)$ and the recursive step function $I\times I \to I \times I$, $(x, y) \mapsto (x, xy)$ to get a continuous function $N \times I \to I \times I$ such that $(s^n(z), x) \mapsto (x, x^n)$ for $x\in I$, $n \in \IN$.
The sequence $(s^n(z)) \in N$ has a convergent subnet $(s^{n_\lambda}(z))_{\lambda \in \Lambda}$, say with limit $y$. Thus, for any $x\in I$ and $\lambda \in \Lambda$, we have $(s^{n_\lambda}(z), x) \mapsto (x, x^{n_\lambda})$.
Taking limits, we see $(y, x) \mapsto (x, 0)$ if $x \ne 1$ or $(y, x) \mapsto (x, 1)$ if $x = 1$. In other words, $(y, x) \mapsto (x, \delta_{x, 1})$ for all $x\in I$. However, that contradicts the fact that the composition
$$I \overset{y \times \id}\longrightarrow N\times I \to I\times I \overset{p_2}\longrightarrow I \\ x \mapsto (y, x) \mapsto (x, \delta_{x,1}) \mapsto \delta_{x,1},$$
would have to be continuous.
special_objects:
terminal object:
description: singleton space
initial object:
description: empty space
special_morphisms:
epimorphisms:
description: surjective continuous maps (which are automatically quotient maps)
reason: For the non-trivial direction, and for a proof of the parenthetical remark, see the proof above that $\CompHaus$ is epi-regular. |
The sample looks great, and I agree with the comments by others that having everything directly related to the category in one file should be easier to work with. |
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Thank you for the feedback! The mentioned PR is a lot of work and I wanted to make sure that I am going into the right direction. Apparently, yes. |
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... but I have decided against using markdown (long story...), so instead of |
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Interesting search... |
| We let $\CompHaus$ denote the category of compact Hausdorff spaces. | ||
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| \begin{lemma} | ||
| The functor $\Hom({-}, [0, 1])) : \CompHaus^{\op} \to \Set$ is monadic. |
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| The functor $\Hom({-}, [0, 1])) : \CompHaus^{\op} \to \Set$ is monadic. | |
| The functor $\Hom({-}, [0, 1]) : \CompHaus^{\op} \to \Set$ is monadic. |
| We use the crude monadicity theorem. First, the functor has a left adjoint $S \mapsto [0, 1]^S$ with the evident isomorphism | ||
| $$\Hom_{\CompHaus}(X, [0, 1]^S) \simeq \Hom_{\Set}(S, \Hom_{\CompHaus}(X, [0, 1])). $$ | ||
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| To see the functor is conservative, suppose we have a continuous function $f : X \to Y$ such that $f^* : \Hom(Y, [0, 1]) \to \Hom(X, [0, 1])$ is a bijection. Then for any $x_1, x_2 \in X$ with $x_1 \ne x_2$, there exists $\varphi : X \to [0, 1]$ with $\varphi(x_1) = 0$ and $\varphi(x_2) = 1$. Since $f^*$ is surjective, there exists $\psi : Y \to [0, 1]$ with $\varphi = \psi \circ f$; thus, we must have $f(x_1) \ne f(x_2)$. Likewise, we know the image of $f$ is closed. If this image is not all of $Y$, then there exists nonzero $\varphi : Y \to [0, 1]$ which is zero on the image. But then $\varphi \circ f = 0 \circ f$, contradicting the injectivity of $f^*$. Thus, $f$ is a bijective continuous function, and therefore a homeomorphism. |
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please cite Urysohn's Lemma and Tietze's theorem where appropriate
| The functor $\Hom({-}, [0, 1])) : \CompHaus^{\op} \to \Set$ is monadic. | ||
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| \begin{proof} | ||
| We use the crude monadicity theorem. First, the functor has a left adjoint $S \mapsto [0, 1]^S$ with the evident isomorphism |
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Let's add a reference that proves the crude monadicity theorem. For some reason, I could not find one. In Barr-Wells' TTT it is just said that it is clear. :D
| 'CompHaus', | ||
| 'locally copresentable', | ||
| TRUE, | ||
| 'A proof can be found <a href="/pdf/comphaus_copresentable.pdf">here</a>.' |
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Please also add an original reference for this result (see my comments). Either here or in the pdf. Otherwise, readers might get a wrong impression with regards to orginality.
In any case, I think it is very good if we offer a well-written proof directly in CatDat. Thank you for adding it!
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| To see the functor is conservative, suppose we have a continuous function $f : X \to Y$ such that $f^* : \Hom(Y, [0, 1]) \to \Hom(X, [0, 1])$ is a bijection. Then for any $x_1, x_2 \in X$ with $x_1 \ne x_2$, there exists $\varphi : X \to [0, 1]$ with $\varphi(x_1) = 0$ and $\varphi(x_2) = 1$. Since $f^*$ is surjective, there exists $\psi : Y \to [0, 1]$ with $\varphi = \psi \circ f$; thus, we must have $f(x_1) \ne f(x_2)$. Likewise, we know the image of $f$ is closed. If this image is not all of $Y$, then there exists nonzero $\varphi : Y \to [0, 1]$ which is zero on the image. But then $\varphi \circ f = 0 \circ f$, contradicting the injectivity of $f^*$. Thus, $f$ is a bijective continuous function, and therefore a homeomorphism. | ||
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| Finally, suppose we have a coreflexive equalizer pair $E \xhookrightarrow{i} A \overset{f}{\underset{g}{\rightrightarrows}} B$ with $r : B \to A$. We may assume that $i$ is a subspace inclusion map. We may use $r$ to think of $B$ as a bundle of compact spaces over $A$, with two sections $f, g$. We then need to show that |
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| Finally, suppose we have a coreflexive equalizer pair $E \xhookrightarrow{i} A \overset{f}{\underset{g}{\rightrightarrows}} B$ with $r : B \to A$. We may assume that $i$ is a subspace inclusion map. We may use $r$ to think of $B$ as a bundle of compact spaces over $A$, with two sections $f, g$. We then need to show that | |
| Finally, suppose we have a coreflexive equalizer pair | |
| $$E \xhookrightarrow{i} A \overset{f}{\underset{g}{\rightrightarrows}} B$$ | |
| with $r : B \to A$. We may assume that $i$ is a subspace inclusion map. We may use $r$ to think of $B$ as a bundle of compact spaces over $A$, with two sections $f, g$. We then need to show that |
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| By compactness of $X$, we may take finitely many such basic open neighborhoods of the form $p_j^{-1}(V)$ which cover $X$. Again using the assumption that $\I$ is cofiltered, we may assume that $j$ is the same for each neighborhood. In particular, we see that whenever we have $x, y\in X$ with $p_j(x) = p_j(y)$, then $|\varphi(x) - \varphi(y)| < 1/n$. | ||
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| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. |
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Typo:
| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. | |
| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|\varphi(x) - \varphi(y)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. |
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| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. | ||
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| Likewise, suppose we have $\varphi, \psi : X_i \to [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. |
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| Likewise, suppose we have $\varphi, \psi : X_i \to [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. | |
| Likewise, suppose we have $\varphi, \psi : X_i \rightrightarrows [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. |
Also, let's denote the maps differently because
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| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. | ||
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| Likewise, suppose we have $\varphi, \psi : X_i \to [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. |
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Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty.
The pdf needs to be self-contained. Imagine someone gives you link to this pdf. Then this reference is not clear.
Suggestion: Move the lemma from 001_lemmas.sql (whose proof I reformulated, btw) to a lemma in this pdf. Of course, then in the proof of cofiltered-limit-stable epimorphisms, refer to the pdf.
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| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. | ||
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| Likewise, suppose we have $\varphi, \psi : X_i \to [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. |
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Using the contrapositive of this, we can conclude that if
$x \notin \im(p_i)$ , then there exists$f : j \to i$ such that$x \not\in \im(X_f)$ .
I don't understand this part.
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| Now choose such a $j_n$ for each $n$, and then use the fact that $\I$ is $\aleph_1$-cofiltered to take a cone $(k, f_n : k \to j_n)$. We then see that whenever we have $x, y\in X$ with $p_k(x) = p_k(y)$, then $\varphi(x) = \varphi(y)$. Thus, $\varphi$ induces a well-defined function on the image of $p_k$. This function is continuous, since by construction for any $n\in \INnz$ and $x \in X$, there is a neighborhood $V$ of $p_k(x)$ such that whenever $p_k(y)\in V$ as well, $|p_k(y) - p_k(x)| < 1/n$. This induced function can then can be extended to a continuous function $\psi : X_k \to [0,1]$ such that $\varphi = \psi \circ p_k$. | ||
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| Likewise, suppose we have $\varphi, \psi : X_i \to [0,1]$ such that $\varphi \circ p_i = \psi \circ p_i$. For each $n\in \INnz$, consider the set $D_n := \{ x \in X_i \mid |\varphi(x) - \psi(x)| \ge 1/n \}$. For any $x \in D_n$, we must have $x \notin \im(p_i)$. Now recall that we previously proved a result that a cofiltered limit of non-empty compact Hausdorff spaces is nonempty. Using the contrapositive of this, we can conclude that if $x \notin \im(p_i)$, then there exists $f : j \to i$ such that $x \not\in \im(X_f)$. Now $D_n \setminus \im(X_f)$ is open and such sets cover $D_n$; taking a finite subcover and then using the cofiltering assumption again, we conclude that there is a single $f_n : j_n \to i$ such that $\im(X_{f_n})$ is disjoint from $D_n$. Using the $\aleph_1$-cofiltering assumption to take a cone of the $f_n$, we get that there is $f : k \to i$ such that $\im(X_f)$ is disjoint from all $D_n$. This means that $\phi \circ F_f = \psi \circ F_f$. |
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Last sentence must be:
This means that $\phi \circ X_f = \psi \circ X_f$.
| \end{lemma} | ||
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| \begin{corollary} | ||
| The category $\CompHaus$ is $\aleph_1$-copresentable. |
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| The category $\CompHaus$ is $\aleph_1$-copresentable. | |
| The category $\CompHaus$ is locally $\aleph_1$-copresentable. |
(as in the title)
| 'CompHaus', | ||
| 'semi-strongly connected', | ||
| TRUE, | ||
| 'Every non-empty compact Hausdorff space is weakly terminal (by using constant maps).' | ||
| ), |
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| 'CompHaus', | |
| 'semi-strongly connected', | |
| TRUE, | |
| 'Every non-empty compact Hausdorff space is weakly terminal (by using constant maps).' | |
| ), | |
| 'CompHaus', | |
| 'semi-strongly connected', | |
| TRUE, | |
| 'This is already true for <a href="/category/Top">$\mathbf{Top}$</a>.' | |
| ), |
maybe?
| 'exact cofiltered limits', | ||
| FALSE, | ||
| 'Consider the $\IN$-codirected systems $X_n := [0, 1] \times [0, 1/n]$ with the maps $X_{n+1} \to X_n$ being inclusion maps, and $Y_n := [0, 1+1/n]$ with the maps $Y_{n+1} \to Y_n$ also being inclusion maps. We define $f_n : X_n \to Y_n$, $(x, y) \mapsto x$ and $g_n : X_n \to Y_n$, $(x, y) \mapsto x+y$. It is straightforward to check these give morphisms of $\IN$-codirected systems in $\CompHaus$.<br> | ||
| Now for each $n$, the coequalizer of $f_n$ and $g_n$ is a single-point space. On the other hand, $\lim X_n \simeq [0, 1] \times \{ 0 \}$; $\lim Y_n \simeq [0, 1]$; and $\lim f_n = \lim g_n$, $(x, 0) \mapsto x$. Thus, the coequalizer of $\lim f_n$ and $\lim g_n$ is $[0, 1]$, showing that this coequalizer is not preserved under limits.' |
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Can wee add the proof that the coequalizer of
We need to prove that every (continuous) map
More generally, let
Case 1: Values on
If
Case 2: Values on
Let
2 participants
Addresses: #156
Currently undecided properties:
has cartesian filtered colimits
is coaccessible
has cofiltered-limit-stable epimorphisms
is coregular
is countably distributive
has exact cofiltered limits
is infinitary distributive
is locally copresentable
has a natural numbers object