This question concerns the exact same question as on this post: Closed subset of a variety is also a variety
More specifically, the person says that showing $(Y, \mathcal{O}_Y)$ is a $k$-space is easy however I am having trouble showing it. In case definitions do not align this is my definition of a $k$-space:
A $k$-space is a tuple $(X, \mathcal{O}_X)$ such that $X$ is a top. space and every open $U\subset X$, $\mathcal{O}_X(U)\subset\{f:U\to k\}$ is a sub-$k$-algebra such that the following holds:
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For all $V\subset U$ opens and $f\in\mathcal{O}_X(U)$, we have $f|_{V}\in\mathcal{O}_X(V)$.
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For all $U\subset X$ open and for all $f:U\to k$, $f\in\mathcal{O}_X(U)$ if and only if for all $P\in U$ there is an open $U_P\subset U$ such that $P\in U_P$ and $f|_{U_P}\in\mathcal{O}_X(U_P)$.
For every open $V\subset Y$ we define $\mathcal{O}_{Y}(V):= \{f:V\rightarrow k\rvert \forall P\in V, \ \exists U\subset X, \ \exists g\in\mathcal{O}_{X}(V) \ \text{s.t} \ P\in U, \forall Q\in V\cap U \ \text{we have} \ f(Q) = g(Q)\}$.
I cannot seem to figure out how to show the right to left implication of property 2. My attempt goes as follows: Let $P\in U$ then there exists a $U_P\subset U$ such that $P\in U_P$ and $f|_{U_P}\in\mathcal{O}_Y(U_P)$. Then by regularity at $P$ there exists an open $V_P\subset X$ and a $g_P\in\mathcal{O}_X(V_P)$ such that $P\in V_P$ and $f=f|_{U_P}=g_P$ on $U_P\cap V_P$. Then we note that $U=\bigcup_{P\in U}U_P$ so define $V:=\bigcup_{P\in U} V_P$ and $g:V\to k$ by $g(Q)=g_P(Q)$ whenever $Q\in V_P$.
Now here I am having trouble showing that $g$ is well-defined (if it even is). If I let $Q\in V_P\cap V_{P'}$ then $f(Q)=g_P(Q)=g_{P'}(Q)$ only if $Q\in U_P\cap U_P'$ as well, so this approach seems unredeemable. I feel like this is the natural thing to do though as I need some open $P\in V'\subset X$ and $g\in\mathcal{O}_X(V')$ such that $f=g$ on $U\cap V'$.
Best Answer
The idea is that being in $\mathcal{O}_Y$ is a local condition, so you don't need to take unions and stuff. The details are not too enlightening, but here they are in any case:
Let $V\subseteq Y$ be open and $f\colon V\to k$ such that for all $P\in V$ there exists $V_P\subseteq V$ open (so $V_P=U_P\cap Y$ for some open $U_P\subseteq X$) such that $f\vert_{V_p}\in\mathcal{O}_Y(V_P)$. By definition of $\mathcal{O}_Y$, for every $Q\in V_P$ there exists an open $W_Q\subseteq X$ containing it and $g_Q\in\mathcal{O}_X(W_Q)$ such that $f\vert_{V_P\cap W_Q}=g_Q\vert_{V_P\cap W_Q}$. Thus for every $P\in V$ we can take $U_P\cap W_P\subseteq X$ and $g_P\vert_{W_P\cap U_P}\in\mathcal{O}_X(W_P\cap U_P)$ to get $f\vert_{V\cap U_P\cap W_P}=g_P\vert_{V\cap U_p\cap W_p}$, implying that $f\in\mathcal{O}_Y(V)$.