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This question relates to my previous question found here: Defining Category of Problems

Let $\left\{\Pi_i \right\}_{i \in I}$ be a family of problems. Let the solution $u_i$ of $\Pi_i$ lie in some space $X_i$. I want to make this family into a category in order to study decomposition of problems into products of smaller ones. I want to define the morphisms of this category in a manner that, the existence of a morphism $f_{j}^i : \Pi_i \rightarrow \Pi_j$ is equivalent to saying that solving $\Pi_i$ implies solving $\Pi_j$.

(from here and onwards re-edited)

I define a morphism $f_{j}^i : \Pi_i \rightarrow \Pi_j$ to be a map $f: X_i \rightarrow X_j$ such that $f(u_i) = u_j$ and $f$ does not depend on $u_j$.

Question: does this definition capture the interpretation of morphisms that i want? I.e. let $f_{j}^i : \Pi_i \rightarrow \Pi_j$ be a morphism. Does it follow that solving $\Pi_i$ implies solving $\Pi_j$?

thanks :-)

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    You can always decide which diagrams you want to use. These diagrams decide which morphisms are available and which are not.2011-11-17
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    Is the problem of defining the morphisms an object in the Category of problems? :-)2011-11-17
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    @tp1: I think i see your point:) Could you please give me a little bit more detail? So what you are saying is that given the family of problems, the arrows are defined automatically. Right?2011-11-17
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    Interesting thoughts! but what is your formal definition of "problem"?2011-11-18
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    @Bruno: Thanks. I don't have a formal definition for what a problem is. I am thinking of a well-defined mathematical problem. Maybe we can think of it as a map from the space of parameters of the problem to the solution space, e.g. solve $\alpha x = \beta \in \mathbb{R}$, i can think of it as map $\mathbb{R}^2 \rightarrow \mathbb{R}$.2011-11-18
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    @Asaf: No, i do not consider this problem inside the given family :-)2011-11-18
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    IMHO you should clarify further what is a problem. $\alpha x=\beta$ has no solution if $\alpha=0\land\beta\neq 0$, has all solutions if $\alpha=0\land\beta=0$; this case does not fit into $\mathbb{R}^2\to\mathbb{R}$.2011-11-18
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    I'm bewildered of indexes used everywhere. Do you really need an indexed family of problems? http://en.wikipedia.org/wiki/Indexed_family Maybe, just a set suffice? Do you need to index solutions? What means that $f$ does not depend on $u_j$?2011-11-18
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    At first glance you defined a category of subsets. (I can give a formal definition if you like.) Because every subset is a 1-ary relation, this is a particular case of a category of relational systems.2011-11-18
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    @beroal: You are right about $\alpha x=\beta$. Now by saying that $f$ does not depend on $u_j$, i mean that in order to apply the map $f$, it is not necessary to know $u_j$. Otherwise, we always have the trivial map $u_i \rightarrow u_j$.2011-11-18
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    I've voted to close because without a more specific definition of "problem", it is virtually impossible to answer this question. Please [edit] to include one or two concrete examples of morphisms that you want to capture; I'd be more than happy to remove my close vote.2015-03-11

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