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I am studying the proof of the following proposition in Hartshorne - Let $k$ be an algebraically closed field. There is a natural fully faithful functor $Var(k)\longrightarrow Sch(k)$ from the category of varieties over $k$ to schemes over $k$.

First they make some statements about general topological spaces (in which I have a doubt).

Let $X$ be any topological space. And let $t(X)$ be the set of non-empty irreducible closed subsets of $X$. If $Y, Y_1, Y_2, Y_i$ $(i\in I)$ are closed subsets of $X$, then we can check that $t(Y)\subset t(X)$, $t(Y_1)\cup t(Y_2)=t(Y_1\cup Y_2)$ and $\bigcap_{i\in I}t(Y_i)=t(\bigcap_{i\in I}Y_i\big)$. From this, we can make $t(X)$ a topological space, by taking closed subsets of $t(X)$ to be subsets of the form $t(Y)$, where $Y$ is a closed subset of $X$. Further if $f:X_1\longrightarrow X_2$ is a continuous, we obtain a map $t(f):t(X_1)\longrightarrow t(X_2)$ by sending $Y\in t(X_1)$ to $cl(f(Y))$ in $X_2$. Thus $t$ is a functor on topological spaces.

Further, we define a map $\alpha:X \longrightarrow t(X)$, by $\alpha(P)=cl\{P\}$. We see that $\alpha$ is continuous because for any closed subset $t(Y)$ of $t(X)$, $\alpha^{-1}(t(Y))=Y$, which is closed in $X$. Next they make the following statement :

$\alpha$ induces a bijection between the set of open subsets of $X$ and the set of open subsets of $t(X)$. This is where I have doubts.

I think it is enough to prove that there is a bijection between the corresponding closed sets. What I have so far is : if $Y$ is a closed subset of $X$, then $\alpha^{-1}(t(Y))=Y$, and $\alpha(Y)\subset t(Y)$. Also if $Y\neq Z$ closed subsets of $X$, I can show that $\alpha(Y)\neq\alpha(Z)$. What more is needed to prove that there is a bijection?

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Let $U=X\setminus Y$ be an open subset of $X$ (of course $Y$ is closed). The induced map can be defined as $$ \alpha(U) = t(X)\setminus t(Y), $$ which is well defined since $\alpha(U)$ is an open subset of $t(X)$ by definition of its topology. This map is obviously a bijection between the open subsets of $X$ and the open subsets of $t(X)$.

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  • $\begingroup$ We already have a map $\alpha:X\longrightarrow t (X) $. Is the $\alpha $ you have defined, induced by it? $\endgroup$ – gradstudent Dec 12 '13 at 18:30
  • $\begingroup$ Yes, this is what I mean. The original $\alpha$ is defined on points, the induced one on open subsets, but they are morally the same map $\endgroup$ – Abramo Dec 12 '13 at 18:36
  • $\begingroup$ Thank you! I understand it now. $\endgroup$ – gradstudent Dec 13 '13 at 2:42
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I'm not sure Abramo's answer above is correct as described here.

To be clear, we can define a bijection from closed sets of $X$ to closed sets of $tX$ by sending a closed set $Y \subseteq X$ to the closed set $tY \subseteq tX$. This map is clearly subjective by definition of the topology on $tX$. As was mentioned above, we have $\alpha^{-1}(tY) = Y$, for any closed $Y \subset X$, and this gives injectivity.

Closed set and open sets are in bijection by complementation so the above bijection can be made to be between open sets by sending an open $X-Y$ in $X$ to $tX - tY$ in $tX$.

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Let me elaborate more on Abramo's answer. In fact we are looking at $\alpha^{-1}$ for the correspondence of open subsets.

Let $Y\subset X$ be closed so that $X-Y$ is open. Consider $\alpha^{-1}(tX-tY) = X-\alpha^{-1}(tY)$.

What is $\alpha^{-1}(tY)$? It is just $Y$: for $P\in X$, $\alpha(P) =\overline{\{P\}}$ is an irreducible subset of $Y$ if and only $P\in Y$! (Note that the closure of a point is always irreducible).

Thus, we get a bijection between open sets of $X$ and open sets of $tX$, by

$$tX-tY \longleftrightarrow X-Y.$$

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