Morphism of schemes $f\colon X\to Y$ associated to a continuous map of the underlying spaces $|X|\to |Y|$ I am sorry for asking two questions in one but they are strongly related. 


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*What is an example of (affine?) schemes $X=(|X|,\mathcal{O}_X)$ and $Y=(|Y|,\mathcal{O}_Y)$ and a map of topological spaces $|f|\colon|X|\to |Y|$ that cannot be promoted into a map $f\colon X\to Y$ of schemes?


I guess something like $exp:\mathbb{R}\to\mathbb{R}$ is an example but I cannot prove that it is an example.


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*What is an example of (affine?) schemes $X=(|X|,\mathcal{O}_X)$ and $Y=(|Y|,\mathcal{O}_Y)$ and a map of topological spaces $|f|\colon|X|\to |Y|$ that can be promoted into a map $f_1\colon X\to Y$ of schemes and into a map $f_2\colon X\to Y$ a map of schemes with $f_1\neq f_2$?

 A: Just take $X=\mathrm{Spec}(K)$ and $Y=\mathrm{Spec}(L)$ for two fields $K,L$.
There is a unique map $|X| \to |Y|$. The morphisms $X \to Y$ correspond to field homomorphisms $L \to K$. There may be no such homomorphisms, but there may be also many of them. (For example, consider $\mathbb{Q}(\sqrt{2}) \to \mathbb{Q}$ or $\mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2})$).
A: 1) Take $X=Y=\mathbb A^1_\mathbb C$, so that $|X|=\mathbb C\sqcup \{\eta\}$.
 Any permutation of $|X|$ fixing the generic point $\eta$ induces a homeomorphism $f:|X|\to |X|$ .
If the induced permutation on the subset of closed points  $\mathbb C\subset |X|$ is not continuous in the classical topology, then the map $f$ cannot come from a scheme morphism $X\to X$.
As an example you can take the permutation exchanging $0$ and $1$ and fixing everything else.  
2) Given a field $k$ and a non-trivial automorphism $\phi:k\to k$, the induced scheme morphisms $\phi^*, Id^*:\text {Spec}(k)\to \text {Spec}(k)$ are different but induce the same homeomorphism (of one point spaces!) $|\text {Spec}(k)|\to |\text {Spec}(k)|$ 
A: An interesting example is afforded by a discrete valuation ring $R$ and its quotient field $k$. $spec \ R=\{(0), \mathfrak{p}\}$, with $\mathfrak{p}$ closed and $(0)$ open. And $spec \ k=\{(0)\}$.
 Now the inclusion map $R\rightarrow k$ induces $spec \  k\rightarrow spec \ R$ where 
$(0)\mapsto (0)$ the map on sheaves is $k\rightarrow k$ which is local.
However $(0) \mapsto \mathfrak{p}$ is also a continous map. However the induced map in sheaves is $R\rightarrow k$, the inclusion map and this is not a local map since the inverse image of $(0)$ is not the maximal ideal of $R$.
