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Let $I$ be the category with objects points of $[0,1]$ with unique morphism for every pair of objects. Let $C$ be a complete category, suppose that there is an isomorphism $m: a \to b$, $m \in C$ when does there exist a functor $F: I \to C$, with $I(0)=a$, $I(1)=b$, and such that the following holds. Let $D(t)$ denote the directed system $$t_1 \mapsto t _{2} \ldots \mapsto t_i \mapsto \ldots $$ for $t_i \in I$, for all $i$, and $t_i$ converges to $t$. Then $colim_{D(t)} F = F(t)$, for every $t$ and every $D(t)$ as above.

Edit: The original version of the question was trivial as answered below. I added the "continuity" assumption in the form of the colimit condition.

Edit2: Simplified the colimit condition and changed $I$.

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  • $\begingroup$ I suspect you might want to strengthen your continuity condition even more. As it stands you can still have the following not-very-continuous-looking situation: However $F$ has been defined on the interval $[0,\tau]$ you can pick any $b$ with a map $m:F(\tau)\to b$ and define $F(t)=b$, $F(\tau\leq t)=m$ for $t>\tau$, with $F(t\leq t')=\mathrm{id}_B$. Your definition imposes an analogue of lower semicontinuity, so maybe you also want to add the dual criterion that $F$ commutes with limits from the right. $\endgroup$ – Oscar Cunningham Apr 16 '17 at 7:23
  • $\begingroup$ You may not have noticed but definition of the category $I$ has been changed. $\endgroup$ – Yasha Apr 17 '17 at 20:03
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This is a bit of a boring answer, but such a functor always exists. We can use the functor that sends $0$ to $a$, every thing else in $[0,1]$ to $b$ and then sends the morphisms to $\mathrm{id}_a$, $\mathrm{id}_b$ or $m$ as appropriate.

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  • $\begingroup$ Yeah, somehow this is not what I expected, perhaps I need to sharpen the question in some way. $\endgroup$ – Yasha Apr 15 '17 at 22:13
  • $\begingroup$ Note that the above does not answer the current question. $\endgroup$ – Yasha Oct 11 '17 at 17:15
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Yes, certainly. For instance, you could define $F(t)=a$ for $a<1$ and $F(1)=b$, with the obvious choice of morphisms (the identity whenever possible, and otherwise $m$).

More generally, if $I=A\cup B$ is any partition of $I$ such that every element of $A$ is less than every element of $B$ (so $A$ and $B$ are (possibly degenerate) intervals), you could define $F(t)=a$ for $t\in A$ and $F(t)=b$ for $t\in B$.

Your continuity condition, however, cannot always be satisfied. For instance, suppose $C$ is the category whose only objects are $a$ and $b$ and whose only non-identity morphism is $m$. Then it is easy to see any $F$ must come from a partition $I=A\cup B$ as above. Your continuity condition then says that $A$ and $B$ are both closed intervals, which is impossible.

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  • $\begingroup$ I added a final paragraph addressing your edit. $\endgroup$ – Eric Wofsey Apr 15 '17 at 22:34
  • $\begingroup$ Uhm that was only semi-continuity :) fixed. $\endgroup$ – Yasha Apr 15 '17 at 23:20
  • $\begingroup$ I have updated the answer again. $\endgroup$ – Eric Wofsey Apr 15 '17 at 23:31
  • $\begingroup$ Yep, I agree. But when does it exist? $\endgroup$ – Yasha Apr 16 '17 at 0:48

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