## Category of infinite ordinals

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### Category of infinite ordinals

Let $\kappa$ be an infinite regular ordinal. Define $\Delta_\kappa$ to be the category of ordinals strictly smaller than $\kappa$ (i.e. the ordinals which are elements of $\kappa$, in von Neumann definition), with order-preserving maps as morphisms.

Question. Suppose $f\colon \alpha \to \beta$ is a monomorphism (resp. an epimorphism) of $\Delta_\kappa$. Is it true that necessarily $\alpha \leqslant \beta$ (resp. $\alpha \geqslant \beta$)?

Remark. It's easy to prove that monomorphisms are precisely injective order-preserving maps (resp. epimorphisms are precisely surjective oder-preserving maps), so let's assume that.
Feanor

Messaggi: 741
Iscritto il: lun 29 set 2008, 20:58

### Re: Category of infinite ordinals

Let start by proving the first part of the question:
Suppose $f \colon \alpha \to \beta$ is a monomorphism of $\Delta_k$ then $\alpha \leq \beta$.

We know that $f$ is injective and order preserving, by that it is easy to prove the claim: let $\gamma$ be the greater ordinal between $\alpha$ and $\beta$, clearly we have the two inclusion $\alpha \hookrightarrow \gamma$ and $\beta \hookrightarrow \gamma$, so we can extend $f$ to a map $f' \colon \alpha \to \gamma$ (by composing $f$ with the embedding of $\beta$ in $\gamma$).

Clearly $f'$ is still monotone and injective, being the composition of two monotone and injective morphisms, and by construction we know that $\alpha \leq \gamma$,
that is $\alpha$ is an initial segment of $\gamma$.

We have to prove that for each $\alpha' \in \alpha$ (i.e. for each $\alpha' < \alpha$) we have that $f'(\alpha') \geq \alpha'$.
We proceed by transfinite induction.

For $\alpha'=0$ the claim is trivally true: $f'(0) \geq 0$ because every ordinal is great or equal to $0$.

Let assume that for every $\alpha'' < \alpha' \in \alpha$ we have that $f'(\alpha'') \geq \alpha''$. Then we have that for each $\alpha''< \alpha'$

$f'(\alpha') > f'(\alpha'') \geq \alpha''$
because of the monotonicity of $f'$ and the inductive hypothesis.

Now we distinguish between two cases:

• if $\alpha'=\alpha''+1$ then we have that $f'(\alpha') > f'(\alpha'') \geq \alpha''$ and so $f'(\alpha') \geq \alpha''+1=\alpha'$
• if $\alpha'$ is a limit ordinal then $f'(\alpha') > \alpha''$ for each $\alpha'' < \alpha'$, hence $f'(\alpha') \geq \sup\{\alpha'' \colon \alpha'' < \alpha'\}=\alpha'$
whatever the case we always end up having $f(\alpha') \geq \alpha'$, as we wished to prove.

By transfinite induction it follows that $f'(\alpha') \geq \alpha'$ for every $\alpha' \in \alpha$.
Since $f'$ has image contained in $\beta$ this proves that for each $\alpha' \in \alpha$ we have that

$\alpha' \leq f'(\alpha') \in \beta$
hence

$\alpha' \in \beta$
for each $\alpha' \in \alpha$, that is $\alpha \leq \beta$ and so by general fact of the ordinal $\alpha \leq \beta$.
Ultima modifica di ineff il mar 27 giu 2017, 16:15, modificato 1 volta in totale.
Con il giusto funtore si può risolvere ogni problema.
ineff

Messaggi: 186
Iscritto il: sab 17 apr 2010, 19:45

### Re: Category of infinite ordinals

Now back to the second part of the question:

If $f \colon \alpha \to \beta$ is an epimorphism in $\Delta_k$ then $\alpha \geq \beta$.

By hypothesis $f$ is an epimorphism, hence surjective and monotone. In particular this implies that for every $\gamma_1,\gamma_2 \in \beta$ and every $\alpha_1,\alpha_2 \in \alpha$ such that

• $\gamma_1 < \gamma_2$
• $f(\alpha_1)=\gamma_1$
• $f(\alpha_2)=\gamma_2$
we have that $\alpha_1 < \alpha_2$: clearly $\alpha_1$ and $\alpha_2$ have to be different having different images via $f$, and if by absurd $\alpha_2 < \alpha_1$ this would imply that

$\gamma_2=f(\alpha_2) \geq f(\alpha_1) =\gamma_1$
against the hypothesis that $\gamma_1 < \gamma_2$.

So we can consider the family of disjoint sets $(f^{-1}(\{\gamma\}))_{\gamma \in \beta}$, whose elements are not empty sets since $f$ is surjective by hypothesis.
By axioms of choice we get a mapping

$g \colon \beta \to \bigcup_{\gamma \in \beta} f^{-1}(\{\gamma\}) \subseteq \alpha$
such that for every $\gamma \in \beta$ we have $g(\gamma) \in f^{-1}(\{\gamma\})$.

Whenever $\gamma_1 < \gamma_2$ we have $f(g(\gamma_1))=\gamma_1$ and $f(g(\gamma_2))=\gamma_2$ and so, by the previous remark, it follows that $g(\gamma_1) < g(\gamma_2)$.

This implies that $g \colon \beta \to \alpha$ is a strictly monotone map, i.e. a monomorphism, that by question 1 implies that $\beta \leq \alpha$.
Con il giusto funtore si può risolvere ogni problema.
ineff

Messaggi: 186
Iscritto il: sab 17 apr 2010, 19:45

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