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?
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$\begingroup$ 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. $\endgroup$– Damiano MazzaNov 5, 2022 at 6:39
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$\begingroup$ 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. $\endgroup$– LaRNov 5, 2022 at 8:32
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$\begingroup$ 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. $\endgroup$– Damiano MazzaNov 5, 2022 at 11:34
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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.
answered Nov 5, 2022 at 12:00