Existence of "Continuous paths" in categories as directed systems 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 $I$.
 A: 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.
A: 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.
