Do continuous functors generalise continuous maps? [duplicate]

In category theory we say a functor is continuous if it preserves limits.

At first glance this seems like a totally obvious analogue of the notion of a continuous mapping in topology. However, at second glance it looks like a play on words, because the category-theoretic notion of "limit of a diagram" doesn't have a lot to do with the topological notion of "limit of a sequence".

Can one regard continuity of maps as a special case of continuity of functors? That is, if I have topological spaces $$X, Y$$ and an arbitrary mapping $$f: X \to Y$$, is there a natural way to cook up a functor $$F_f$$ between some categories such that $$F_f$$ is a continuous functor iff $$f$$ is a continuous map?

• Category-theoretic limit related to topological limit?
– ZxJx
Oct 29 '19 at 20:46
• "Continuous" means "commutes with limits." In Damien L's answer, he shows that the two notions of limits coincide. I think the important part to note is that, when one says "continuous function," they mean in the topological category. Now extending this to functors...
– ZxJx
Oct 29 '19 at 21:02

One way to connect those notions is to consider any functor $$F : J \rightarrow C$$, where $$J$$ is a category consisting of one object ("X") and one non-identity endomorphism $$f : X \rightarrow X$$. Then a cone of F (chosen among all cones using index i) satisfies $$Ff \circ h_X = h_X$$, where $$h_X : A_i \rightarrow FX$$, so each cone is a fixed point of $$Ff$$. Therefore a limit of $$F$$ is the universal cone with unique mediating morphism $$m_i : A_i \rightarrow L$$ for each cone in such way that the choice of cone doesn't impact $$L$$. In other words, L is the least fixed point of $$Ff$$, since L is also (a vertex of) a cone of F. Since $$Ff : FX \rightarrow FX$$ can be any endomorphism in C by choice of F, the categorical limit allows producing least fixed point of any endomorphisms, with "least" defined by least element in the ordering imposed by the endomorphisms in C. In particular, you can construct natural numbers by choosing $$F(X) = N$$ and $$Ff(x) = x+1$$. $$L=\infty$$ in that case represents the limit of all natural numbers (the least infinite number). Now given a sequence $$s : N \rightarrow A$$ in C that preserves limits, we can find an element for all natural numbers, but because $$s$$ preserves limits, also one for their limit. So $$s_{\infty}$$ is defined and must correspond to a limit of the sequence defined by $$s$$.