Suppose $(X,O_X)$ and $(Y, O_Y)$ are ringed spaces, and let $X = U_1 \cup U_2$ be an open cover. Suppose we have morphism of ringed spaces $\pi_i: U_i \rightarrow Y$ such that they agree on the overlap. I want to show that there is a unique morphism of ringed spaces $\pi: X \rightarrow Y$ such that $\pi|_{U_i} = \pi_i$.

I was able to show that the continuous map between $X$ and $Y$, but I am having trouble defining the map between the structure sheaves. I appreciate any help! Thank you!

PS I would also appreciate if someone could explain me exactly what it means to "agree on the overlaps", what does this mean for the map of sheaves of the ringed space morphism?


1 Answer 1


I am learning the same topics. I will try to answer this.

Two morphism agree on the overlap means they map the open subsets of the overlap to the same subsets in $Y$ and also they preserve the regular functions: $\pi_1^* \phi = \pi_2^* \phi$ on any open subsets of the overlap, for any regular function $\phi$ on $Y$, restricted on $\pi_i(U_1\cap U_2)$, where $\pi_i^* \phi =\phi \circ \pi_i$.

Define the morphism $\pi=\begin{cases}\pi_1 \text{ on } U_1\\ \pi_2 \text{ on } U_2\end{cases}$

You have proved the continuity, so now it remains to prove for any open subset $V\subset Y$, $\phi\in O_Y(V)$ implies $\pi^*\phi \in O_X(\pi^{-1}(V))$.

We know that $(\pi|_{U_i\cap\pi^{-1}(V)})^*\phi \in O_X(U_i\cap\pi^{-1}(V))$ since $\pi|_{U_i\cap\pi^{-1}(V)}$ is a morphism. And $\pi^*\phi$ agrees on the intersection of $U_i\cap\pi^{-1}(V)$. By the gluing property for sheaves, $\pi^*\phi\in O_X(\pi^{-1}(V))$.

Hope this makes sense to you.

  • $\begingroup$ Thank you for your answer! I am having trouble understanding what you mean by $\phi \pi_i$. I am a bit confused because $\phi \in O_Y(V)$ for some open set $V$ and $\pi_i$ is a map from $U_i$ to $Y$. $\endgroup$
    – user211392
    Mar 4, 2015 at 1:25
  • $\begingroup$ $\phi \pi_i$ maps $\pi_i(V^{-1}\cap U_i)$ to $K$. It is the pull-back of $\phi$ by $\pi_i$. Because $\pi_i$ maps $\pi_i(V^{-1}\cap U_i)$ back to $V$ again, then $\phi$ maps it to $K$. So $\phi \pi_i$ becomes a regular function in $O_X(V^{-1}\cap U_i)$. $\endgroup$
    – KittyL
    Mar 4, 2015 at 9:54

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