<p>The answer is no.</p>
<p>Let me first rewrite the definition given in the comments: an arrow <span class="math-container">$f:D\to E$</span> of <span class="math-container">$\mathbf{CPO}$</span> is an <em>embedding</em> if there exists <span class="math-container">$g:E\to D$</span> such that <span class="math-container">$g\circ f=\mathrm{id}_D$</span> and <span class="math-container">$f\circ g\sqsubseteq\mathrm{id}_E$</span>, where <span class="math-container">$\sqsubseteq$</span> denotes the pointwise order.</p>
<p>Observe that, being right-invertible, embeddings are monic, so the arrows of <span class="math-container">$\mathbf{CPO}^E$</span> are all monic.  Now, for any category <span class="math-container">$\mathcal C$</span> with a terminal object, if all terminal arrows are monic, then <span class="math-container">$\mathcal C$</span> is posetal (there is at most one arrow between any two objects): let <span class="math-container">$1$</span> be the terminal object, let <span class="math-container">$f,g:A\to B$</span> be two parallel arrows and let <span class="math-container">$t:B\to 1$</span> be the unique terminal arrow from <span class="math-container">$B$</span>; we have <span class="math-container">$t\circ f=t\circ g$</span> (because <span class="math-container">$1$</span> is terminal), so <span class="math-container">$f=g$</span> because <span class="math-container">$t$</span> is a monomorphism.</p>
<p>Now, it is immediate to see that <span class="math-container">$\mathbf{CPO}^E$</span> is not posetal: let <span class="math-container">$\mathrm{Unit}$</span> be the cpo <span class="math-container">$\{\bot,\top\}$</span> and let <span class="math-container">$\mathrm{Bool}$</span> be the cpo <span class="math-container">$\{\bot,0,1\}$</span> with <span class="math-container">$0$</span> and <span class="math-container">$1$</span> incomparable; there are two distinct embeddings <span class="math-container">$\mathrm{Unit}\to\mathrm{Bool}$</span>.  Therefore, <span class="math-container">$\mathbf{CPO}^E$</span> cannot have a terminal object.</p>

CC BY-SA 4.0
2022-11-05T12:00:08.683Z
2022-11-05T17:29:25.993Z
2022-11-05T17:29:25.993Z
Damiano Mazza
<p>Let <span class="math-container">${\bf CPO}$</span> be the category of <span class="math-container">$\omega$</span>-complete partial orders and <span class="math-container">$\omega$</span>-continuous functions. Let <span class="math-container">${\bf CPO}^{E}$</span> be the category of embeddings of <span class="math-container">${\bf CPO}$</span>. Does <span class="math-container">${\bf CPO}^{E}$</span> have a terminal object?</p>

CC BY-SA 4.0
2022-11-05T06:39:23.673Z
I don't know what you mean by "embedding" but as long as 1) the unique map $\{\bot,\top\}\to\{\bot\}$ is not an embedding, and 2) any map $f:\{\bot,\top\}\to X$ such that $f(\top)\neq\bot$ is an embedding, then the answer is "no", and these seem to me to be pretty reasonable requirements.
Damiano Mazza
CC BY-SA 4.0
2022-11-05T08:32:38.797Z
Copying from Tennent's book: "If $D$ and $E$ are domains, a continuous function $f: D\rightarrow E$ is an embedding just if there is a continuous partial function $g:E \rightarrow D$ (termed a projection) such that: (1) $g \cdot f = id_D$, and (2) $f \cdot g \sqsubseteq id_E$". So, I think, your second condition is not always satisfied.
LaR
CC BY-SA 4.0
2022-11-05T11:34:47.083Z
Ok, so these are "open" embeddings.  The answer is still no, because they are monomorphisms, and the only way for terminal arrows to be monos is that the category is posetal, which is not the case of $\mathbf{CPO}^E$.  I'll write up an answer.
Damiano Mazza
<semantics>
Terminal object in the category of embeddings
LaR
# Terminal object in the category of embeddings

(Asked by: username_0 on 2022-11-04)

Let ${\bf CPO}$ be the category of $\omega$-complete partial orders and $\omega$-continuous functions. Let ${\bf CPO}^{E}$ be the category of embeddings of ${\bf CPO}$. Does ${\bf CPO}^{E}$ have a terminal object?

**Tags:** semantics

> **Comments:**
> **username_1**: I don't know what you mean by "embedding" but as long as 1) the unique map $\{\bot,\top\}\to\{\bot\}$ is not an embedding, and 2) any map $f:\{\bot,\top\}\to X$ such that $f(\top)\neq\bot$ is an embedding, then the answer is "no", and these seem to me to be pretty reasonable requirements.  
> **username_0**: Copying from Tennent's book: "If $D$ and $E$ are domains, a continuous function $f: D\rightarrow E$ is an embedding just if there is a continuous partial function $g:E \rightarrow D$ (termed a projection) such that: (1) $g \cdot f = id_D$, and (2) $f \cdot g \sqsubseteq id_E$". So, I think, your second condition is not always satisfied.  
> **username_1**: Ok, so these are "open" embeddings.  The answer is still no, because they are monomorphisms, and the only way for terminal arrows to be monos is that the category is posetal, which is not the case of $\mathbf{CPO}^E$.  I'll write up an answer.  

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## 1 answers

### Answer by username_1 on 2022-11-05. Score: 6 | Accepted answer

The answer is no.

Let me first rewrite the definition given in the comments: an arrow $f:D\to E$ of $\mathbf{CPO}$ is an *embedding* if there exists $g:E\to D$ such that $g\circ f=\mathrm{id}\_D$ and $f\circ g\sqsubseteq\mathrm{id}\_E$, where $\sqsubseteq$ denotes the pointwise order.

Observe that, being right-invertible, embeddings are monic, so the arrows of $\mathbf{CPO}^E$ are all monic. Now, for any category $\mathcal C$ with a terminal object, if all terminal arrows are monic, then $\mathcal C$ is posetal (there is at most one arrow between any two objects): let $1$ be the terminal object, let $f,g:A\to B$ be two parallel arrows and let $t:B\to 1$ be the unique terminal arrow from $B$; we have $t\circ f=t\circ g$ (because $1$ is terminal), so $f=g$ because $t$ is a monomorphism.

Now, it is immediate to see that $\mathbf{CPO}^E$ is not posetal: let $\mathrm{Unit}$ be the cpo $\{\bot,\top\}$ and let $\mathrm{Bool}$ be the cpo $\{\bot,0,1\}$ with $0$ and $1$ incomparable; there are two distinct embeddings $\mathrm{Unit}\to\mathrm{Bool}$. Therefore, $\mathbf{CPO}^E$ cannot have a terminal object.


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