Add natural numbers objects without parameters

content/natural_numbers_objects.md

@@ -0,0 +1,103 @@
+---
+title: Natural numbers objects
+description: We prove some results on natural numbers objects.
+---
+
+## Natural numbers objects
+
+The definition of a [natural numbers object](/category-property/natural_numbers_object) a priori only allows for recursively defined morphisms in which the next value $\Phi(s(n))$ depends only on the previous value $\Phi(n)$. In many cases, however, we would also like to use $n$ itself to define $\Phi(s(n))$. This can be done in categories with finite products:
+
+::: Lemma 1
+Let $(N,z,s)$ be a natural numbers object in a category with finite products. Then for every $a : 1 \to X$ and every $g : N \times X \to X$ there exists a unique morphism $\Phi : N \to X$ such that $\Phi(z) = a$ and $\Phi(s(n)) = g(n, \Phi(n))$.
+:::
+
+Here, we use element notation to simplify the exposition. For example, the equation $\Phi(s(n)) = g(n,\Phi(n))$ means that the following diagram commutes:
+
+$$
+\begin{CD}
+N @>{s}>> N \\
+@V{(\id_N,\Phi)}VV @VV{\Phi}V \\
+N \times X @>>{g}> X
+\end{CD}
+$$
+
+_Proof._ Define the morphism $b : 1 \to N \times X$ by $b \coloneqq (z,a)$ and the morphism $h : N \times X \to N \times X$ by $h(n,x) \coloneqq (n,g(n,x))$. By the universal property of $(N,z,s)$, there is a unique morphism $\Psi : N \to N \times X$ such that:
+
+- $\Psi(z) = b$
+- $\Psi(s(n)) = h(\Psi(n))$
+
+Write $\Psi(n) = (\Psi_0(n),\Psi_1(n))$, where $\Psi_0 : N \to N$ and $\Psi_1 : N \to X$. The two equations above then become:
+
+- $\Psi_0(z) = z$
+- $\Psi_0(s(n)) = \Psi_0(n)$
+- $\Psi_1(z) = a$
+- $\Psi_1(s(n)) = g(\Psi_0(n),\Psi_1(n))$
+
+The uniqueness in the universal property of $(N,z,s)$ implies $\Psi_0 = \id_N$. Therefore, $\Phi \coloneqq \Psi_1$ is the unique morphism $\Phi : N \to X$ satisfying $\Phi(z)=a$ and $\Phi(s(n)) = g(n,\Phi(n))$. <span class="qed">$\square$</span>
+
+The next result appears in [Johnstone](https://ncatlab.org/nlab/show/Sketches+of+an+Elephant), Part A, Lemma 2.5.5. Our proof is slightly more concise because we have extracted Lemma 1.
+
+::: Lemma 2
+Let $(N,z,s)$ be a natural numbers object in a category with finite products. Then
+$$1 \xrightarrow{z} N \xleftarrow{s} N$$
+is a coproduct cocone. Thus, $N \cong 1 \sqcup N$.
+:::
+
+_Proof._ Let $a : 1 \to X$ and $b : N \to X$ be morphisms. We need to show that there is a unique morphism $c : N \to X$ satisfying $c(z) = a$ and $c(s(n)) = b(n)$. This follows by applying Lemma 1 to the morphism $g : N \times X \to X$ defined by $g(n,x) \coloneqq b(n)$. <span class="qed">$\square$</span>
+
+Next, we will check when the terminal object $1$ itself is a natural numbers object. In that case, $z : 1 \to 1$ and $s : 1 \to 1$ are necessarily equal to $\id_1$.
+
+::: Lemma 3
+Let $1$ be a terminal object in a category. Then $(1,\id_1,\id_1)$ is a natural numbers object if and only if for every endomorphism $g : X \to X$ and every morphism $a : 1 \to X$ we have $g \circ a = a$. If the category has finite products, $(1,\id_1,\id_1)$ is a parametrized natural numbers object if and only if $g = \id_X$ for every endomorphism $g : X \to X$, i.e. the category is [one-way](/category-property/one-way).
+:::
+
+_Proof._ The first statement is immediate from the definition of a natural numbers object. For the second, $(1,\id_1,\id_1)$ is a parametrized natural numbers object if and only if, for all morphisms $f : A \to X$ and all endomorphisms $g : X \to X$, there is a unique morphism $\Phi : A \to X$ such that $\Phi \circ \id_A = f$ and $\Phi \circ \id_A = g \circ \Phi$. These equations simplify to $\Phi = f$ and $f = g \circ f$. Since this must hold for every $f : A \to X$, we must have $g = \id_X$ (by the Yoneda Lemma or by a direct argument). <span class="qed">$\square$</span>
+
+Next, we prove a partial converse to the [result](/category-implication/parametrized_nno_criterion) that countably distributive categories have a parametrized natural numbers object.
+
+::: Lemma 4
+Let $\C$ be a category with finite products, countable copowers denoted $\otimes$, and a parametrized natural numbers object $1 \xrightarrow{z} N \xrightarrow{s} N$. Then there is an isomorphism $N \cong \IN \otimes 1$, and for every object $A$ the natural morphism
+$$\alpha : \IN \otimes A \to A \times (\IN \otimes 1)$$
+is an isomorphism.
+:::
+
+_Proof._ We will use element notation extensively. In particular, for every element $a \in A$ and $n \in \IN$, there is an element $n \otimes a \in \IN \otimes A$, formally defined by the $n$th coproduct inclusion. The morphism $\alpha$ is defined by
+$$\alpha(n \otimes a) = (a,n \otimes 1).$$
+
+In any category with a terminal object and countable copowers, we can construct the non-parametrized NNO $\IN \otimes 1$ with the element $0 \otimes 1 \in \IN \otimes 1$ and the map
+$$s : \IN \otimes 1 \to \IN \otimes 1, \quad s(n \otimes 1) \coloneqq (n+1) \otimes 1.$$
+See [here](/category-implication/nno_criterion) for a detailed proof. Since, by assumption, $1 \xrightarrow{z} N \xrightarrow{s} N$ is a parametrized NNO, it is also a non-parametrized NNO and is therefore isomorphic to the one just constructed. We may assume without loss of generality that they are equal and hence work with $N = \IN \otimes 1$.
+
+Next, apply the parametrized universal property of the NNO to the diagram
+$$A \xrightarrow{f} \IN \otimes A \xrightarrow{g} \IN \otimes A$$
+defined by $f(a) \coloneqq 0 \otimes a$ and $g(n \otimes a) \coloneqq (n+1) \otimes a$. It gives a morphism
+$$\Phi : A \times N \to \IN \otimes A$$
+satisfying
+$$\Phi(a,0 \otimes 1) = 0 \otimes a, \quad \Phi(a,s(m)) = g(\Phi(a,m)).$$
+For $m \coloneqq n \otimes 1 \in N$, where $n \in \IN$, the second equation becomes
+$$\Phi(a,(n+1) \otimes 1) = g(\Phi(a,n \otimes 1)).$$
+By induction on $n \in \IN$, it follows that
+$$\Phi(a,n \otimes 1) = n \otimes a,$$
+which is exactly the statement that $\Phi \circ \alpha = \id_{\IN \otimes A}$.
+
+It remains to prove $\alpha \circ \Phi = \id_{A \times N}$. We first observe that
+$$\alpha \circ g = (\id_A \times s) \circ \alpha$$
+as morphisms $\IN \otimes A \to A \times (\IN \otimes 1)$. Indeed, for every $n \otimes a \in \IN \otimes A$,
+$$\alpha(g(n \otimes a)) = \alpha((n+1) \otimes a) = (a, (n+1) \otimes 1),$$
+while
+$$(\id_A \times s)(\alpha(n \otimes a)) = (\id_A \times s)(a, n \otimes 1) = (a, (n+1) \otimes 1).$$
+The universal property applied to the diagram
+$$A \xrightarrow{(\id_A,z)} A \times N \xrightarrow{\id_A \times s} A \times N$$
+shows that there is a _unique_ morphism $\Psi : A \times N \to A \times N$ satisfying
+$$\Psi \circ (\id_A,z) = (\id_A,z)$$
+and
+$$\Psi \circ (\id_A \times s) = (\id_A \times s) \circ \Psi,$$
+namely $\id_{A \times N}$. Thus, it suffices to verify that $\alpha \circ \Phi : A \times N \to A \times N$ satisfies these two equations. We have
+$$\alpha(\Phi(a,z)) = \alpha(0 \otimes a) = (a, 0 \otimes 1) = (a,z),$$
+and hence $\alpha \circ \Phi \circ (\id_A,z) = (\id_A,z)$. Moreover,
+$$\alpha \circ \Phi \circ (\id_A \times s) = \alpha \circ g \circ \Phi = (\id_A \times s) \circ \alpha \circ \Phi.$$
+This finishes the proof. <span class="qed">$\square$</span>
+
+_Remark._ Actually, the mentioned [result](/category-implication/parametrized_nno_criterion) and Lemma 4 can be combined into an equivalent characterization as follows: In a category with finite products and countable copowers, the NNO (which exists, see [here](/category-implication/nno_criterion)) is a parametrized NNO if and only if for all objects $A$ the canonical morphism
+$$\textstyle \coprod_{n \in \IN} A = \coprod_{n \in \IN} (A \times 1) \to A \times \coprod_{n \in \IN} 1$$
+is an isomorphism. This is the precise connection to countable distributivity.

database/data/categories/CompHaus.yaml

@@ -73,9 +73,9 @@ unsatisfied_properties:
     references:
       - haus_no_regular_subobject_classifier
 
-  - property: natural numbers object
+  - property: parametrized natural numbers object
     proof: >-
-      Let $I \coloneqq [0, 1]$. 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
+      Let $I \coloneqq [0, 1]$. If a parametrized 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
       $$\begin{align*}
       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},

database/data/categories/Delta.yaml

@@ -57,12 +57,12 @@ satisfied_properties:
     proof: The category $\FinOrd \setminus \{\varnothing\}$ is core-thin because already <a href="/category/FinOrd">$\FinOrd$</a> is core-thin.
 
   - property: mono-regular
-    proof: The proof for <a href="/category/FinOrd">$\FinOrd$</a> also works for $\FinSet \setminus \{\varnothing\}$.
+    proof: The proof for <a href="/category/FinOrd">$\FinOrd$</a> also works for $\FinOrd \setminus \{\varnothing\}$.
     references:
       - finord_mono-regular
 
   - property: epi-regular
-    proof: The proof for <a href="/category/FinOrd">$\FinOrd$</a> also works for $\FinSet \setminus \{\varnothing\}$.
+    proof: The proof for <a href="/category/FinOrd">$\FinOrd$</a> also works for $\FinOrd \setminus \{\varnothing\}$.
     references:
       - finord_epi-regular
 
@@ -81,9 +81,6 @@ satisfied_properties:
       - finset_aleph1-cofiltered-limits
 
 unsatisfied_properties:
-  - property: strict terminal object
-    proof: This is trivial.
-
   - property: cofiltered
     proof: 'The two maps $d^0,d^1 : [0] \rightrightarrows [1]$ are not equalized by any morphism.'
 
@@ -100,6 +97,9 @@ unsatisfied_properties:
     references:
       - finord_no_sequential_limits
 
+  - property: natural numbers object
+    proof: Any natural numbers object in $\FinOrd \setminus \{\varnothing\}$ would also be a natural numbers object in <a href="/category/FinOrd">$\FinOrd$</a>, which we know does not exist.
+
   - property: pushouts
     proof: Assume that the two inclusions $\{0 < 1\} \leftarrow \{0\} \rightarrow \{0 < 2\}$ have a pushout in $\FinOrd \setminus \{\varnothing\}$. This would be a universal non-empty finite ordered set $X$ with three elements $0,1,2$ satisfying $0 \leq 1$ and $0 \leq 2$. Assume w.l.o.g. $1 \leq 2$ (the case $2 \leq 1$ is similar). The universal property yields an order-preserving map $X \to \{a < b < c\}$ with $0 \mapsto a$, $1 \mapsto c$, $2 \mapsto b$. But then $c \leq b$, which is a contradiction.
 

database/data/categories/FinOrd.yaml

@@ -71,15 +71,9 @@ unsatisfied_properties:
   - property: skeletal
     proof: This is trivial.
 
-  - property: one-way
-    proof: There are three different order-preserving maps $\{0 < 1\} \to \{0 < 1\}$.
-
   - property: countable
     proof: This is trivial.
 
-  - property: strict terminal object
-    proof: This is trivial.
-
   - property: sequential limits
     proof: Consider the (non-empty) ordered set $[n] \coloneqq \{0 < \cdots < n\}$ for $n \in \IN$. The forgetful functor to $\Set$ is representable, hence preserves all limits. Thus, if the diagram of truncation maps $\cdots \twoheadrightarrow [2] \twoheadrightarrow [1] \twoheadrightarrow [0]$ has a limit in $\FinOrd$, its underlying set is isomorphic to the limit taken in $\Set$, which is $\IN \cup \{\infty\}$. But this is not a finite set.
     label: finord_no_sequential_limits
@@ -88,6 +82,9 @@ unsatisfied_properties:
     proof: 'Consider the (non-empty) ordered set $[n] \coloneqq \{0 < \cdots < n\}$ for $n \in \IN$. Assume the sequence of inclusion maps $[0] \hookrightarrow [1] \hookrightarrow [2] \hookrightarrow \cdots$ has a colimit $(f_n : [n] \to X)$ in $\FinOrd$. Let $n_0 \geq 0$ be fixed. I claim that $f_{n_0}$ is injective, which will then yield a contradiction by taking $n_0 \geq \card(X)$. For $n \geq 0$ define $g_n : [n] \to [n_0]$ as follows. For $n \leq n_0$ it is the inclusion, and for $n \geq n_0$ it is the surjection which keeps all elements of $[n_0]$ and maps all other elements to $n_0$. Observe that $g_n$ preserves the order and $g_{n+1} |_{[n]} = g_n$. Hence, there is a unique order-preserving map $g : X \to [n_0]$ with $g \circ f_n = g_n$ for all $n$. For $n = n_0$ this shows $g \circ f_{n_0} = \id_{[n_0]}$, and $f_{n_0}$ is injective.'
     label: finord_no_sequential_colimits
 
+  - property: natural numbers object
+    proof: 'Assume that there is a natural numbers object $(N,z,s)$, where $z \in N$ and $s : N \to N$. Let $d \coloneqq \card(N)$. Consider the finite ordered set $X \coloneqq \{1 < \dotsc < d+1\}$ and the order-preserving map $g : X \to X$ defined by $g(k) \coloneqq k + 1$ for $k \leq d$ and $g(d+1) \coloneqq d+1$. By the universal property of $(N,z,s)$, there is a unique order-preserving map $\Phi : N \to X$ such that $\Phi(z) = 1$ and $\Phi(s(n)) = g(\Phi(n))$ for all $n \in N$. We then have $\Phi(s(z)) = 2$, $\Phi(s^2(z)) = 3$, and so on. Formally, an induction shows that $\Phi(s^n(z)) = n+1$ for every $0 \leq n \leq d$. Thus, $\Phi$ is surjective, contradicting $\card(N) < \card(X)$.'
+
 special_objects:
   initial object:
     description: empty ordered set

database/data/categories/FinSet.yaml

@@ -55,10 +55,7 @@ unsatisfied_properties:
     proof: This is trivial.
 
   - property: natural numbers object
-    proof: >-
-      If $(N,z,s)$ is a natural numbers object, then
-      $$1 \xrightarrow{z} N \xleftarrow{s} N$$
-      is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no finite set $N$ with $N \cong 1 + N$.
+    proof: By Lemma 2 <a href="/content/natural_numbers_objects">here</a>, if $(N,z,s)$ is a natural numbers object, then $N \cong 1 \sqcup N$. But there is no finite set with this property.
 
 special_objects:
   initial object:

database/data/categories/Met.yaml

@@ -123,9 +123,6 @@ unsatisfied_properties:
   - property: locally finite
     proof: This is obvious.
 
-  - property: strict terminal object
-    proof: This is trivial.
-
   - property: balanced
     proof: The inclusion $\IQ \hookrightarrow \IR$ is a counterexample; it is an epimorphism since $\IQ$ is dense in $\IR$. Alternatively, consider the identity map $(X,2d) \to (X,d)$ for any non-trivial metric space $(X,d)$.
 
@@ -148,10 +145,7 @@ unsatisfied_properties:
     label: met_no_filtered_colimit_stable_monos
 
   - property: natural numbers object
-    proof: >-
-      If $(N,z,s)$ is a natural numbers object in $\Met$, then
-      $$1 \xrightarrow{z} N \xleftarrow{s} N$$
-      is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.
+    proof: By Lemma 2 <a href="/content/natural_numbers_objects">here</a>, if $(N,z,s)$ is a natural numbers object, then $N \cong 1 \sqcup N$. Since there is a map $1 \to N$, we have $N \neq \varnothing$. However, the coproduct of two non-empty metric spaces does not exist, see <a href="https://math.stackexchange.com/questions/1778408" target="_blank">MSE/1778408</a>.
 
   - property: effective congruences
     proof: 'Any kernel pair of $h : X \to Z$ in $\Met$ corresponds to a closed subset of $X\times X$. However, there are plenty of non-closed congruences, such as $\Delta \cup (\IQ \times \IQ) \subseteq \IR \times \IR$ with the subspace metric.'