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Call a ringed space local it if it lies in the image of the obvious faithful, non-full functor from locally ringed spaces to ringed spaces.

Given a ringed space, is there a map $f$ from it to some local ringed space such that any other map from it to a local ringed space much factor uniquely through $f$?

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So you are asking if the forgetful functor from locally ringed spaces to ringed spaces has a left adjoint. See here

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    $\begingroup$ @kartop_man As it happens with the universal property which characterizes every left adjoint, two maps are morphisms in the domain of the left adjoint functor -in this case ringed spaces - and the unique map is a morphism in the codomain, i.e. locally ringed spaces. This is how I understood the question. $\endgroup$
    – display llvll
    Commented Jun 1, 2019 at 16:49
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    $\begingroup$ Thanks to GLe for point out to the MSE question and the paper arxiv.org/pdf/1103.2139.pdf.. Surprisinglt there is some similarity between what I have written before and what is discussed in the paper.. $\endgroup$
    – Praphulla Koushik
    Commented Jun 1, 2019 at 17:18
  • $\begingroup$ I am sorry, GLe was actually right and I indeed was wrong. Her/his answer does in fact completely address the question. My mistake. Maybe the only use from my comments was that I emphasized the fact that the OP's local ringed spaces were not the same as locally ringed spaces in terms of morphisms, which it seems was not noticed by everyone. The end result was the same either way. $\endgroup$
    – user140765
    Commented Jun 1, 2019 at 21:07
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Let $\text{RS}$ be the category of ringed spaces, $\text{LRS}$ the category of locally ringed spaces. As you allude to, there is a forgetful functor $\mathcal{F}$: $\text{LRS} \rightarrow \text{RS}$, that is not full. As GLe notes, this functor has no left adjoint, since it does not preserve limits. An example is provided in the link GLe gave. From that example, I think you should be able to concoct an example of a fiber product (pullback), that isn't preserved.

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  • $\begingroup$ Sure - let $X \in \text{RS}$. Let $g: X \rightarrow Y$, where $Y \in \mathcal{F}(\text{LRS})$. By the condition OP states, there exists a unique map $f$, and a local ringed space $\mathcal{C}$, such that $g$ factors through $f$. All these maps are in the category of ringed spaces. But this is precisely the property that defines a left adjoint for $\mathcal{F}$. $\endgroup$
    – John Rached
    Commented Jun 1, 2019 at 20:57
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This is too long for a comment. So, I am writing here.

Let $(X,\mathcal{O}_X)$ be a ringed space. The idea is to localize the ring $\mathcal{O}_x$ for each open $x\in X$. Let $\mathcal{L}_x$ denote the localization of $\mathcal{O}_x$. Now one can ask if this collection $\{\mathcal{L}_x\}_{x\in X}$ form a sheaf on $X$??

Is this what you are thinking??

The idea is to localize the rings $\mathcal{O}_x$ but there is no obvious choice of prime ideals where we can localize. So, I think we need to fix a prime ideal $P_x$ in $\mathcal{O}_x$ for each $x\in X$. Then, we localize $\mathcal{O}_x$ at $P_x$ and call this $\mathcal{L}_x$. I am almost sure this collection $\{\mathcal{L}_x\}_{x\in X}$ gives a sheaf on $X$.

Universal property will then be depending on the choice of prime ideals $\{P_x:x\in X\}$.

The paper Localization of Ringed spaces by W. D. Gillam is relevant here. I will add more details soon.

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  • $\begingroup$ +1: obvious to whom? This answer-comment may require more work to be helpful. $\endgroup$
    – user140765
    Commented Jun 1, 2019 at 16:32
  • $\begingroup$ I do not understand what you mean by “obvious to whom”.. what does it mean? I agree that this needs little more work to be helpful :) I am trying to add... :) $\endgroup$
    – Praphulla Koushik
    Commented Jun 1, 2019 at 16:33
  • $\begingroup$ I might as well ask you what the phrase in the bold means. One has to define "obvious". $\endgroup$
    – user140765
    Commented Jun 1, 2019 at 16:37
  • $\begingroup$ To define localization, we need a prime ideal.. but a ringed space does not come up with any such collection.. so, I am saying we must be also given a collection of prime ideals.. @kartop_man is this convincing? $\endgroup$
    – Praphulla Koushik
    Commented Jun 1, 2019 at 16:40
  • $\begingroup$ Thanks to GLe for point out to the MSE question and the paper arxiv.org/pdf/1103.2139.pdf.. Surprisinglt there is some similarity between what I have written before and what is discussed in the paper.. $\endgroup$
    – Praphulla Koushik
    Commented Jun 1, 2019 at 17:18

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