Simple Question about Category theory. I'm graduate student studying algebraic topology.
 I'm have a simple Question.
Q. Is there a mathematical notion of "Topology on Category" which can describe the notion of "limits in Category, as like the classical case ?
 A: As I stated in the comments, the problem is usually that such topologies do not distinguish between cones and universal cones. However, in the special case where the categories are preorders (i.e., categories where each $Hom(A, B)$ has at most one arrow), then it is possible to give a topology on the set of objects in such a way that for complete preorders, a functor is continuous with respect to the topologies considered if and only if it preserves limits (i.e., inf) of totally (pre)ordered diagrams. Let us call a functor between complete preorders weakly continuous if it does exactly that, i.e., if it preserves all limits corresponding to totally (pre)ordered diagrams. We can topologize the preorders as follows:
Consider first the category $\mathbf{2}$ consisting of an arrow $0 \to 1$ between two objects. Define a topology there which has the object $1$ as the non-trivial closed set. For a general preorder $\mathcal{C}$, define now a topology as the one whose closed sets are of the form $F^{-1}(1)$ for weakly continuous functors $F: \mathcal{C} \to \mathbf{2}$. To see that it is indeed a topology, note that if $(A_i)_i$ are closed sets corresponding to the preimages of $1$ of the weakly continuous functors $(F_i)_i$, respectively, then $\prod_i F_i$ is a weakly continuous functor such that the preimage of $1$ is exactly $\cap_i A_i$. Similarly, if $A, B$ are closed sets corresponding to the preimages of $1$ of the weakly continuous functors $F, G$, respectively, then $F \coprod G$ is weakly continuous and the preimage of $1$ is exactly $A \cup B$ (note that in general the coproduct of two limit preserving functors from preorders to $\mathbf{2}$ need not be limit preserving, so we do need to restrict our considerations to weakly continuous functors).
It is clear that if $F: \mathcal{C} \to \mathcal{D}$ is weakly continuous, it is continuous with respect to the topologies defined above, since for a closed set $A$ in $\mathcal{D}$, preimage of $1$ of the functor $G$, we have that $F^{-1}(A)$ is the preimage of $1$ of the weakly continuous composition $GF$. Conversely, let us see that if $F$ is continuous with respect to the topologies then it must be weakly continuous. If $(C \to C_i)_i$ is a limiting cone in $\mathcal{C}$ corresponding to a totally (pre)ordered diagram $(C_i)_i$, then by definition $C$ belongs to the closure of $(C_i)_i$, and hence $F(C)$ must belong to the closure of $(F(C_i))_i$ in $\mathcal{D}$. If $(F(C) \to F(C_i))_i$ were not a universal cone, let $D$ be the vertex of such a cone; we have an induced arrow $F(C) \to D$ (and hence no arrow $D \to F(C)$). But then the representable functor $[D, -]$ (regarded as a functor with values in $\mathbf{2}$) would be weakly continuous (in fact, limit preserving), and the closed set which is the preimage of $1$ contains each $F(C_i)$ but not $F(C)$, which is absurd, since in that case $F(C)$ would not belong to such a closed set containing the $F(C_i)$. Therefore, $(F(C) \to F(C_i))_i$ must be a universal cone and the proof is complete.
