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Topology would not "work" if one reverted the "direction" in the definition of continuous maps $f$:

$$\text{open}(x) \rightarrow \text{open}(f(x))$$

It has to be

$$\text{open}(f(x)) \rightarrow \text{open}(x)$$

For graphs – among others – things look different. You can equally define graph homomorphisms as mappings $f$ satisfying

$$\text{R}(x,y) \rightarrow \text{R}(f(x),f(y))$$

or satisfying

$$\neg\text{R}(x,y) \rightarrow \neg\text{R}(f(x),f(y))$$

which is equivalent with

$$\text{R}(f(x),f(y)) \rightarrow \text{R}(x,y) $$

What is the lesson to be learned from this observation? What distinguishes topological spaces from graphs (with their respective "natural" morphisms)?

Is there another – maybe more categorical – formulation of this observation?

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Why do you say that graph homomorphisms could be defined as maps satisfying $$\neg\text{R}(x,y) \rightarrow \neg\text{R}(f(x),f(y))$$ That certainly is not what I would conceive of as a graph homomorphism, but I have not studied graph theory in any depth. –  Zev Chonoles Jul 22 '13 at 22:34
    
I just guessed: graph homomorphisms could be defined that way - graph theory and any category of graphs would not be affected seriously. –  Hans Stricker Jul 22 '13 at 22:50
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@HansStricker: You would definitely get a different notion of graph homomorphism, hence a different category. Just like "open map" is different from "continuous map" so you would get a different category of topological spaces if you took those to be the morphisms. So why are you ok with changing the category of graphs and their homomorphisms, but not ok with changing the category of topological spaces? –  Jim Jul 22 '13 at 23:29
    
@Jim: Because I believe to have understood that "open maps" don't yield as interesting results as "continuous maps" - other than in the case of graphs, where the "positive" and the "negative" morphisms yield - essentially - the same results. But I may be mislead. –  Hans Stricker Jul 22 '13 at 23:42
    
I don't know what $\mathrm{open}(x) \to \mathrm{open}(f(x))$ means, but if you think the analogue of what you did for graphs is taking open maps between topological spaces, that's wrong: the analogue is taking maps of spaces $f : X \to Y$ such that whenever $f^{-1}(U)$ is open, $U$ itself is open. –  Omar Antolín-Camarena Jul 23 '13 at 14:40
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1 Answer

Let's first have a look at graphs with edge-preserving maps and graphs with edge-reflecting maps. Given some graph $R$, we have the dual graph $\neg R$ which has edges exactly where $R$ has no edges. Then $f : R \to Q$ is edge-preserving iff $f : \neg R \to \neg Q$ is edge-reflecting. So, while we do have two different categories here, they are equivalent. Edge-preserving maps are exactly as interesting as edge-reflecting maps (for arbitrary graphs).

There is no corresponding self-dual in topology. If we exchange the open sets with the non-opens, we don't get a new topology. So continuous maps and open maps may differ. The reason why the continuous maps are the interesting ones is the existence of Sierpinski space $\mathbb{S} := (\{\bot,\top\}, \{\emptyset, \{\top\}, \{\bot,\top\})$. A subset of a topological space is open iff its characteristic function into $\mathbb{S}$ is continuous. The open functions don't have a similar way to recover the open sets. In fact, you can do topology by starting with the continuous functions and $\mathbb{S}$, and then deriving the rest (called synthetic topology).

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