Add total/cototal category properties - #254
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I have some rough ideas on some of the others: on Hausdorff spaces and semigroups, I think I should be able to use an idea similar to the one for Cat to keep control over the images of constant maps. For CMon, I think the "subdirectly irreducible" property might have to do with limiting the number of maps to it - though I'm not yet at all sure how to translate that into a contradiction. And on locally ringed spaces, I have a vague idea that I might be able to define a functor whose L(T) would have a number of maps from Spec k which grows faster than possible for any single locally ringed space. Anyway, no rush on reviewing this - I was just working on this off and on over the past week, and wanted to get the progress so far pushed before resuming work on the quasitopos PR. |
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It would be cool if we can merge this soon. The few remaining cases don't have to be dealt with at this moment. |
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OK, what does have to be resolved before we can merge it: There are several places where references are missing or incomplete. And the current proof that total -> complete needs to be finished, or replaced with a reference if for some reason finishing off the proof that the construction does give a limit is too complex. |
Oups I forgot that the proofs are incomplete. |
Co-authored-by: Script Raccoon <scriptraccoon@gmail.com>
Also update definitions to versions that make sense even for non locally small categories
…yper-cocomplete" to adapt the terminology in Kelly
dschepler
changed the title
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@ScriptRaccoon I think this PR is ready for review now. |
Great! I will have a look in the next days. |
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Based on the diff in this file and the content page file in the other commit: clearly, your editor uses different formatting settings than mine. I use VS Code and the Prettier extension; it is also listed in the file extensions.json. The formatting settings are defined in the file .prettierrc. What is your setup? Can we find a way that we format the files in the same way?
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I've been using Qt Creator.
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Can you create a configuration file that your editor understands?
| - initial object | ||
| proof: 'The empty diagram is vacuously a discrete fibration; thus, it has a colimit, which must be an initial object.' | ||
| - initial object | ||
| proof: The empty diagram is vacuously a discrete fibration; thus, it has a colimit, which must be an initial object. |
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Notice that the quotes for single line strings are only necessary when the text has some YAML-special characters, typically :
This is why I have removed them here.
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I have done quite a bit of rewording here. Please check if this is OK for you. I am afraid that the git diff is useless.
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I have the cspell VS code extension which underlines unknown or incorrect words. For unknown / mathematical words I then click quick fix > add to config. Can you perhaps also use it?
The extension is useful since it catches lots of typos. (For example, it caught your typo cocotal in content/missing_cogenerator.md that I fixed in the other commit.)
| Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. <span class="qed">$\square$</span> | ||
| Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. | ||
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| Now assume that $\C$ is locally essentially small and cototal. Using the axiom of choice, we may assume that for each small cardinal $\kappa$, there is at most one element $X \in \F$ such that $\card(U(X)) = \kappa$. Treating $\F$ as a discrete diagram in $\C$, assumption (1) implies that for any object $Y$ of $\C$, the collection of cocones $\F \to Y$ is bijective with a set, since the maps $X \to Y$ with $\card(U(X)) > \card(U(Y))$ must all be zero in such a cocone. Therefore, by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, $\C$ must have a coproduct $Y$ of all elements of $\F$. But then by assumption (2), there exists $X \in \F$ such that $\card(U(X)) > \card(U(Y))$; and since $\C$ is pointed, the coprojection $X \to Y$ must be split monic and therefore non-zero. Using assumption (1), we get a contradiction. <span class="qed">$\square$</span> |
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Using the axiom of choice, we may assume that for each small cardinal
$\kappa$ , there is at most one element$X \in \F$ such that$\card(U(X)) = \kappa$ .
Why? If we discard elements from
| Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. <span class="qed">$\square$</span> | ||
| Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. | ||
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| Now assume that $\C$ is locally essentially small and cototal. Using the axiom of choice, we may assume that for each small cardinal $\kappa$, there is at most one element $X \in \F$ such that $\card(U(X)) = \kappa$. Treating $\F$ as a discrete diagram in $\C$, assumption (1) implies that for any object $Y$ of $\C$, the collection of cocones $\F \to Y$ is bijective with a set, since the maps $X \to Y$ with $\card(U(X)) > \card(U(Y))$ must all be zero in such a cocone. Therefore, by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, $\C$ must have a coproduct $Y$ of all elements of $\F$. But then by assumption (2), there exists $X \in \F$ such that $\card(U(X)) > \card(U(Y))$; and since $\C$ is pointed, the coprojection $X \to Y$ must be split monic and therefore non-zero. Using assumption (1), we get a contradiction. <span class="qed">$\square$</span> |
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Please explain in which way (the dual of) Thm. 5.6 applies. It says cototal => cocompact => hypercocomplete => ..., where does it say that diagrams with "few cocones" have colimits?
| - property: cogenerating set | ||
| proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of commutative $R$-algebras which are fields: If $F$ is a commutative $R$-algebra that is also a field and $A$ is a non-trivial commutative $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' | ||
| - property: cototal | ||
| proof: 'Let $\F$ be the family of commutative $R$-algebras of the form $R \times k$ where $k$ is an infinite field including a quotient field of $R$. Then for any commutative $R$-algebra $A$, we have a distinguished morphism $R \times k \to A$ consisting of the projection to $R$ followed by the unique morphism $R \to A$. Moreover, if we have any morphism $\varphi : R \times k \to A$ which is not equal to the distinguished morphism, that implies that $\varphi(0, 1) \ne 0$, so the rng homomorphism $k \to R \times k \to A$ is injective, implying $\card(U(A)) \ge \card(U(k))$. From here, an argument similar to the one <a href="/content/missing_cogenerator">here</a> gives a contradiction, using the distinguished morphisms in place of zero morphisms.' |
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The lemma has been added to unify various proofs. If we only write now "an argument similar to ...", the lemma has lost its purpose. I suggest to either write down a self-contained proof or find a variant of the lemma that handles this. Maybe (a spontaneous guess!) also a reduction to the category of non-unital commutative algebras is possible (which is pointed): you are working with augmented algebras here, which are equivalent to non-unital algebras.
| - property: cogenerating set | ||
| proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of $R$-algebras which are fields: If $F$ is an $R$-algebra that is also a field and $A$ is a non-trivial $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' | ||
| - property: cototal | ||
| proof: Essentially the same proof as for <a href="/category/CAlg(R)">$\CAlg(R)$</a> works here. |
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Try to find a variant of the lemma that takes care of both categories.
| - property: cogenerating set | ||
| proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' | ||
| - property: cototal | ||
| proof: 'This is a special case of the proof for <a href="/category/Alg(R)">$\Alg(R)$</a> with $R = \IZ$.' |
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This is a bit inconvenient for the reader since the proof for Alg(R) also redirects to the proof for CAlg(R). (And, as mentioned, this proof is also not fully self-contained right now.)
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Is there any source that defines total categories for categories that are not locally small? Kelly works with enriched categories, and
I find it a bit unfortunate that the current definition in CatDat (1) differs from the one you find in most texts, in particular nlab, and that the left adjoint definition is only mentioned afterwards (2) is quite technical, (3) is presented by two equivalent conditions, both of which look a bit "random" to me.
Do we perhaps want to add "locally essentially small" as a condition?
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Can you please add a bit of context and motivation here in the introduction? Roughly:
- Grp is total by result XYZ
- More generally, every category that is monadic over Set is total by result XYZ
- Nevertheless, we add a proof for Grp here to highlight XYZ and explain XYZ
- The construction of the left adjoint shows in particular that ...
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| - property: cototal | ||
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. |
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Now consider the ultra-wide pushout diagram
$1 \rightrightarrows B S_\kappa$
Please explain this a bit more. I assume you mean the diagram that consists of the unique morphisms
The notation $1 \rightrightarrows B S_\kappa$ is a bit weird since $1 \rightrightarrows B S_\kappa \to \C is a bit weird.
Also, this is not a pushout diagram (it lacks the pushout object).
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Maybe one can use a notation like $X \mathrel{\substack{\rightarrow\\[-0.6ex]\cdots\\[-0.6ex]\rightarrow}} Y$, a variant of the \rightrightrightarrows macro that I have added recently.
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| - property: cototal | ||
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. |
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Maybe it's better to denote this group by
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. | ||
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| On the other hand, we claim that $1 \rightrightarrows B S_\kappa$ does not have a pushout in $\Cat$; by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, this will imply that $\Cat$ is not cototal. To see this, suppose we have a pushout $\C$ of $1 \rightrightarrows B S_\kappa$, and choose a cardinal $\lambda > \card(\Mor(\C))$. Then the coprojection $i_\lambda : B S_\lambda \to \C$ must be split monic, since we can construct a cocone $1 \rightrightarrows B S_\kappa \to B S_\lambda$ in which $B S_\kappa \to B S_\lambda$ corresponds to the zero map for $\kappa \ne \lambda$, and in which $B S_\lambda \to B S_\lambda$ is the identity. It follows that if $X$ is the image in $\C$ of the object of $B S_\lambda$ under $i_\lambda$, then $i_\lambda$ induces an injective map $S_\lambda \to \End_{\C}(X)$. This gives a contradiction since $\lambda > \card(\End_{\C}(X))$ and $S_\lambda$ is a simple group. |
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As in my other comment, please explain in which way Thm. 5.6. gives us that the pushout would exist.
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. | ||
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| On the other hand, we claim that $1 \rightrightarrows B S_\kappa$ does not have a pushout in $\Cat$; by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, this will imply that $\Cat$ is not cototal. To see this, suppose we have a pushout $\C$ of $1 \rightrightarrows B S_\kappa$, and choose a cardinal $\lambda > \card(\Mor(\C))$. Then the coprojection $i_\lambda : B S_\lambda \to \C$ must be split monic, since we can construct a cocone $1 \rightrightarrows B S_\kappa \to B S_\lambda$ in which $B S_\kappa \to B S_\lambda$ corresponds to the zero map for $\kappa \ne \lambda$, and in which $B S_\lambda \to B S_\lambda$ is the identity. It follows that if $X$ is the image in $\C$ of the object of $B S_\lambda$ under $i_\lambda$, then $i_\lambda$ induces an injective map $S_\lambda \to \End_{\C}(X)$. This gives a contradiction since $\lambda > \card(\End_{\C}(X))$ and $S_\lambda$ is a simple group. |
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$\lambda > \card(\Mor(\C))$
Maybe one can briefly mention that this is possible since
| is not injective. Therefore, $\Sub_{\reg} : \Cat^{\op} \to \Set^+$ does not preserve pullbacks, so it cannot be representable. | ||
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| - property: cototal | ||
| proof: >- |
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I think we can make the proof more conceptual by doing something like: if Cat is cototal, then also 1 / Cat is cototal by result XYZ, and then (adjunction?) also Grp is cototal, which is not true. In fact, you are basically repeating the proof that Grp is not cototal here.
Alternatively, can we use the lemma?
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| - property: cototal | ||
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. |
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Do we maybe want to write Aut(X) instead of End(X)?
| whose underlying sets are $\{p\}$ and $\{p,q\}$, respectively. Then $X$ represents the functor sending a semigroup $A$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $A$ that $ab = b$, $ba = a$. It is easy to check that this defines an equivalence relation (see <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for details). Since $p \ne q$ in $E$, the equalizer of the two maps $X \rightrightarrows E$ is the empty semigroup. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of non-empty words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$. | ||
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| - property: cototal | ||
| proof: >- |
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Most of my comments from Cat.yaml also apply here.
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| - property: cototal | ||
| proof: >- | ||
| The proof is similar to the proof for <a href="/category/Cat">$\Cat$</a>. For each infinite cardinal $\kappa$, let $S_\kappa$ be a simple group of cardinality $\kappa$ (such as the alternating group on $\kappa$). We can then form the ultra-wide pushout diagram $1 \rightrightarrows S_\kappa$ in $\SemiGrp$. For every semigroup $A$, the collection of cocones $1 \rightrightarrows S_\kappa \to A$ is bijective to a set: for every such cocone, we must first choose an idempotent $e$ of $A$ corresponding to the map $1 \to A$. Then, whenever $\kappa > \card(U(A))$, then for $f_\kappa : S_\kappa \to A$ in the cocone, we see |
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Explain briefly that
| proof: >- | ||
| The proof is similar to the proof for <a href="/category/Cat">$\Cat$</a>. For each infinite cardinal $\kappa$, let $S_\kappa$ be a simple group of cardinality $\kappa$ (such as the alternating group on $\kappa$). We can then form the ultra-wide pushout diagram $1 \rightrightarrows S_\kappa$ in $\SemiGrp$. For every semigroup $A$, the collection of cocones $1 \rightrightarrows S_\kappa \to A$ is bijective to a set: for every such cocone, we must first choose an idempotent $e$ of $A$ corresponding to the map $1 \to A$. Then, whenever $\kappa > \card(U(A))$, then for $f_\kappa : S_\kappa \to A$ in the cocone, we see | ||
| $$N \coloneqq \{g \in S_\kappa : f_\kappa(g) = e\}$$ | ||
| is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $a$. |
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| is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $a$. | |
| is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $e$. |
| $$N \coloneqq \{g \in S_\kappa : f_\kappa(g) = e\}$$ | ||
| is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $a$. | ||
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| We now claim that $1 \rightrightarrows S_\kappa$ does not have a pushout in $\SemiGrp$; by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, this will imply that $\SemiGrp$ is not cototal. To see this, suppose we had a pushout $A$, and let $\lambda$ be a cardinal strictly greater than $\card(U(A))$. Then the coprojection $S_\lambda\to A$ must be split monic, since we can construct a cocone $1 \rightrightarrows S_\kappa \to S_\lambda$ such that the map $S_\kappa \to S_\lambda$ is the constant map with image 1 if $\kappa \ne \lambda$, while the map $S_\lambda \to S_\lambda$ is the identity. But this contradicts the choice of $\lambda$. |
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Such a pushout would yield a coproduct of the
| $$N \coloneqq \{g \in S_\kappa : f_\kappa(g) = e\}$$ | ||
| is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $a$. | ||
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| We now claim that $1 \rightrightarrows S_\kappa$ does not have a pushout in $\SemiGrp$; by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, this will imply that $\SemiGrp$ is not cototal. To see this, suppose we had a pushout $A$, and let $\lambda$ be a cardinal strictly greater than $\card(U(A))$. Then the coprojection $S_\lambda\to A$ must be split monic, since we can construct a cocone $1 \rightrightarrows S_\kappa \to S_\lambda$ such that the map $S_\kappa \to S_\lambda$ is the constant map with image 1 if $\kappa \ne \lambda$, while the map $S_\lambda \to S_\lambda$ is the identity. But this contradicts the choice of $\lambda$. |
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"is the constant map with image 1" <-- in the proof for Cat, this was called the zero map. Let's make it consistent. I would call it the trivial homomorphism.
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| - property: cototal | ||
| proof: >- | ||
| For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. |
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for example the group of permutations of $\kappa$ of finite support which are even
Maybe formulate it just like in the proof for SemiGrp:
such as the alternating group on $\kappa$
I have also used the term "alternating group" (on infinite sets) in other proofs. Since the sign only makes sense for permutations with finite support, this condition should be clear.
| - property: cototal | ||
| # cspell: disable-next-line | ||
| proof: >- | ||
| For each small cardinal $\kappa$, let $Q_\kappa$ be the product of all Hausdorff topological spaces whose underlying set is a non-empty subset of $\kappa$. By a theorem of Herrlich (<i>Wann sind alle stetigen Abbildungen in Y konstant</i>. Math. Z. 90 (1965): 152-154. <a href="http://eudml.org/doc/170472" target="_blank">EUMDL</a>), there is a regular Hausdorff space $X_\kappa$ with at least two points such that every continuous map $X_\kappa \to Q_\kappa$ is constant. (The author only states that $X_\kappa$ is regular, but actually, $X_\kappa$ is regular and $T_1$, hence Hausdorff.) Choose a base point $x_\kappa \in X_\kappa$ for each $\kappa$. We can form an ultra-wide pushout diagram $1 \rightrightarrows X_\kappa$ where each morphism $1 \to X_\kappa$ corresponds to $x_\kappa$. Then for any Hausdorff space $Y$, the collection of cocones $1 \rightrightarrows X_\kappa \to Y$ is bijective to a set: if $Y$ is empty, then the collection of cocones is obviously empty. Otherwise, in order to form a cocone, we must first choose $y \in Y$ corresponding to the morphism $1 \to Y$. Then for each $\kappa \ge \card(U(Y))$, $Y$ is homeomorphic to one of the spaces in the product forming $Q_\kappa$. Therefore, there is a morphism $Y \to Q_\kappa$ splitting the projection map $Q_\kappa \to Y$. It follows that the map $X_\kappa \to Y$ is constant, and in fact it must be the constant map with image $y$. |
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The citation looks as if the paper proves something about this specific space
| - property: cototal | ||
| # cspell: disable-next-line | ||
| proof: >- | ||
| For each small cardinal $\kappa$, let $Q_\kappa$ be the product of all Hausdorff topological spaces whose underlying set is a non-empty subset of $\kappa$. By a theorem of Herrlich (<i>Wann sind alle stetigen Abbildungen in Y konstant</i>. Math. Z. 90 (1965): 152-154. <a href="http://eudml.org/doc/170472" target="_blank">EUMDL</a>), there is a regular Hausdorff space $X_\kappa$ with at least two points such that every continuous map $X_\kappa \to Q_\kappa$ is constant. (The author only states that $X_\kappa$ is regular, but actually, $X_\kappa$ is regular and $T_1$, hence Hausdorff.) Choose a base point $x_\kappa \in X_\kappa$ for each $\kappa$. We can form an ultra-wide pushout diagram $1 \rightrightarrows X_\kappa$ where each morphism $1 \to X_\kappa$ corresponds to $x_\kappa$. Then for any Hausdorff space $Y$, the collection of cocones $1 \rightrightarrows X_\kappa \to Y$ is bijective to a set: if $Y$ is empty, then the collection of cocones is obviously empty. Otherwise, in order to form a cocone, we must first choose $y \in Y$ corresponding to the morphism $1 \to Y$. Then for each $\kappa \ge \card(U(Y))$, $Y$ is homeomorphic to one of the spaces in the product forming $Q_\kappa$. Therefore, there is a morphism $Y \to Q_\kappa$ splitting the projection map $Q_\kappa \to Y$. It follows that the map $X_\kappa \to Y$ is constant, and in fact it must be the constant map with image $y$. |
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We can form an ultra-wide pushout diagram $1 \rightrightarrows X_\kappa$
See my remarks on Cat.yaml
| proof: >- | ||
| For each small cardinal $\kappa$, let $Q_\kappa$ be the product of all Hausdorff topological spaces whose underlying set is a non-empty subset of $\kappa$. By a theorem of Herrlich (<i>Wann sind alle stetigen Abbildungen in Y konstant</i>. Math. Z. 90 (1965): 152-154. <a href="http://eudml.org/doc/170472" target="_blank">EUMDL</a>), there is a regular Hausdorff space $X_\kappa$ with at least two points such that every continuous map $X_\kappa \to Q_\kappa$ is constant. (The author only states that $X_\kappa$ is regular, but actually, $X_\kappa$ is regular and $T_1$, hence Hausdorff.) Choose a base point $x_\kappa \in X_\kappa$ for each $\kappa$. We can form an ultra-wide pushout diagram $1 \rightrightarrows X_\kappa$ where each morphism $1 \to X_\kappa$ corresponds to $x_\kappa$. Then for any Hausdorff space $Y$, the collection of cocones $1 \rightrightarrows X_\kappa \to Y$ is bijective to a set: if $Y$ is empty, then the collection of cocones is obviously empty. Otherwise, in order to form a cocone, we must first choose $y \in Y$ corresponding to the morphism $1 \to Y$. Then for each $\kappa \ge \card(U(Y))$, $Y$ is homeomorphic to one of the spaces in the product forming $Q_\kappa$. Therefore, there is a morphism $Y \to Q_\kappa$ splitting the projection map $Q_\kappa \to Y$. It follows that the map $X_\kappa \to Y$ is constant, and in fact it must be the constant map with image $y$. | ||
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| On the other hand, we claim that $1 \rightrightarrows X_\kappa$ does not have a pushout in $\Haus$; by G. M. Kelly, <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">A survey of totality for enriched and ordinary categories</a>, Thm. 5.6, this will imply that $\Cat$ is not cototal. To see this, suppose we had a pushout $Y$, and let $\lambda \coloneqq \card(U(Y))$. Then the coprojection $X_\lambda \to Y$ is split monic, since we can construct a cocone $1 \rightrightarrows X_\kappa \to X_\lambda$ where the map $X_\kappa \to X_\lambda$ is the constant map with image $x_\lambda$ when $\kappa \ne \lambda$, and the map $X_\lambda \to X_\lambda$ is the identity. But similarly to the previous paragraph, we can show any morphism $X_\lambda \to Y$ must be constant, giving a contradiction since $X_\lambda$ has at least two points. |
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this will imply that
$\Cat$ is not cototal
typo
2 participants
Unknown categories decided for "total" property:
category of Z-functors
Unknown categories for "cototal" property:
category of commutative monoids
category of locally ringed spaces
category of Z-functors