Add category Mono of injective set functions and commutative squares#266
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Add category Mono of injective set functions and commutative squares#266dschepler wants to merge 1 commit into
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ScriptRaccoon
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Jul 5, 2026
| \end{CD}$$ | ||
| description: >- | ||
| This is the full subcategory of $\Set^{\rightarrow}$ consisting of the injective set functions. It is also equivalent to the category of pairs of a set $X$ and a subset $S \subseteq X$, with morphisms $(X, S) \to (Y, T)$ being the functions $f : X \to Y$ such that $f(S) \subseteq T$. | ||
| nlab_link: https://ncatlab.org/nlab/show/Mono |
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The link doesn't seem to fit.
| proof: This is easy. | ||
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| - property: semi-strongly connected | ||
| proof: This is immediate from the fact that $\Mono$ is a full subcategory of $\Set^{\rightarrow}$, and the latter is semi-strongly connected. |
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Please add a link to the category page. This way, the category and hence the proof can be found without scrolling anywhere else first.
<a href="/category/Set_arrow">$\Set^{\rightarrow}$</a>
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| - property: cogenerator | ||
| proof: >- | ||
| The object $\{ \top, \bot \} \xhookrightarrow{\id} \{ \top, \bot \}$. To see this, suppose we have two morphisms $(\ell_1, r_1), (\ell_2, r_2) : (X, Y, f) \rightrightarrows (X', Y', g)$ such that for every morphism $(\ell', r') : (X', Y', g) \to (\{ \top, \bot \}, \{ \top, \bot \})$, $(\ell', r') \circ (\ell_1, r_1) = (\ell', r') \circ (\ell_2, r_2)$. Then morphisms $(X', Y', g) \to (\{ \top, \bot \}, \{ \top, \bot \}, \id)$ are equivalent to functions $Y' \to \{ \top, \bot \}$. Since $\{ \top, \bot \}$ is a cogenerator in $\Set$, this implies $r_1 = r_2$; and since $g$ is a monomorphism in $\Set$, we automatically get $\ell_1 = \ell_2$ as well. |
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The first sentence seems to be incomplete.
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| - property: generator | ||
| proof: >- | ||
| The object $0 \hookrightarrow 1$ is a generator. This is because it represents the codomain functor taking an object $f : X \hookrightarrow Y$ to $Y$, and taking a morphism $(\ell, r)$ to $r$. That functor is faithful because once $r$ is fixed, $\ell$ is uniquely determined since $f'$ is a monomorphism in $\Set$. |
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Please add what f' is. I think one can write (l,r) : f \to f'
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| - property: cogenerator | ||
| proof: >- | ||
| The object $\{ \top, \bot \} \xhookrightarrow{\id} \{ \top, \bot \}$. To see this, suppose we have two morphisms $(\ell_1, r_1), (\ell_2, r_2) : (X, Y, f) \rightrightarrows (X', Y', g)$ such that for every morphism $(\ell', r') : (X', Y', g) \to (\{ \top, \bot \}, \{ \top, \bot \})$, $(\ell', r') \circ (\ell_1, r_1) = (\ell', r') \circ (\ell_2, r_2)$. Then morphisms $(X', Y', g) \to (\{ \top, \bot \}, \{ \top, \bot \}, \id)$ are equivalent to functions $Y' \to \{ \top, \bot \}$. Since $\{ \top, \bot \}$ is a cogenerator in $\Set$, this implies $r_1 = r_2$; and since $g$ is a monomorphism in $\Set$, we automatically get $\ell_1 = \ell_2$ as well. |
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Isn't this an adjunction argument? Can we reuse something?
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| - property: co-Malcev | ||
| proof: >- | ||
| For this proof we will work in the equivalent category of pairs $(X, X')$ where $X' \subseteq X$. Thus, suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is surjective. Since $\Set$ is co-Malcev, it follows that $E \simeq X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $r(i_1(x)) = x \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case. |
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Let's add a link to the page for Set.
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| - property: co-Malcev | ||
| proof: >- | ||
| For this proof we will work in the equivalent category of pairs $(X, X')$ where $X' \subseteq X$. Thus, suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is surjective. Since $\Set$ is co-Malcev, it follows that $E \simeq X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $r(i_1(x)) = x \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case. |
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Let's use \cong for isomorphism, not \simeq.
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| - property: co-Malcev | ||
| proof: >- | ||
| For this proof we will work in the equivalent category of pairs $(X, X')$ where $X' \subseteq X$. Thus, suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is surjective. Since $\Set$ is co-Malcev, it follows that $E \simeq X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $r(i_1(x)) = x \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case. |
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On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$.
Maybe we can add something like any element of $E'$ is an element of $E$ and hence ...
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| - property: co-Malcev | ||
| proof: >- | ||
| For this proof we will work in the equivalent category of pairs $(X, X')$ where $X' \subseteq X$. Thus, suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is surjective. Since $\Set$ is co-Malcev, it follows that $E \simeq X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $r(i_1(x)) = x \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case. |
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`In the first case, we must have
$r(i_1(x)) = x \in X'$ ``
When I read a chain of relations like this
So here:
| For this proof we will work in the equivalent category of pairs $(X, X')$ where $X' \subseteq X$. Thus, suppose we have a coreflexive corelation $p : (X \sqcup X, X' \sqcup X') \twoheadrightarrow (E, E')$ with coreflexivity morphism $r : (E, E') \to (X, X')$. From the assumption that $p$ is an epimorphism, we have that $p : X \sqcup X \to E$ is surjective. Since $\Set$ is co-Malcev, it follows that $E \simeq X \sqcup_Y X$ for some subset $Y \subseteq X$. It remains to show that $E' = i_1(X') \cup i_2(X') \subseteq X \sqcup_Y X$. Certainly, since we have a morphism $(X \sqcup_Y X, i_1(X') \cup i_2(X')) \to (E, E')$ induced by $p$, we must have $i_1(X') \cup i_2(X') \subseteq E'$. On the other hand, any element of $E'$ is equal to either $i_1(x)$ or $i_2(x)$ for $x \in X$. In the first case, we must have $r(i_1(x)) = x \in X'$, so $i_1(x) \in i_1(X')$; and similarly for the second case. | ||
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| - property: effective cocongruences | ||
| proof: 'See the proof that $\Mono$ is co-Malcev: It shows that in fact any coreflexive corelation is equivalent to an effective cocongruence $X \sqcup X \to X \sqcup_Y X$.' |
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Do you think it's worth adding a property "every reflexive relation is effective" to the database? (Not in this PR of course.)
I think we have seen this a couple of times already (Haus, ...).
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The purpose here is primarily to provide a simple example of a quasitopos which is neither an elementary topos nor thin. (Of course, a lot of the manual proofs here will go away once we can add a "Grothendieck quasitopos" property; this category is equivalent to the$\lnot\lnot$ -separated presheaves on the walking morphism category, where the $\lnot\lnot$ topology is generated by the single morphism $0 \to 1$ forming a covering sieve of $1$ .)