# Why is the induced sheaf on an irreducible closed subset a sheaf?

Supposing that $(X,O_X)$ is a prevariety and that Y is an irreducible closed subset, why is it that for $U\subset Y$ open in the subspace topology, the induced sheaf $O_Y$ with sections $O_Y(U)$ :=[the ring of $k$ valued functions g on U so that for every point $y\in U$ has a neighborhood $V\subset X$ along with a section $G\in O_X(V)$ so that g and G coincide on $U\cap V$] "is" a sheaf on Y. More precisely, I'm failing to see how one can use the ambient sheaf $O_X$ to show that we do indeed obtain a sheaf?

EDIT: After some thought, it would appear to me that there is no need to use the fact that $O_X$ satisfies the sheaf axioms to prove the sheaf axioms for $O_Y$. Am I right in thinking that since being a regular function is a local property then the functions in $O_Y(U)$ are locally regular because they are locally restrictions of regular functions coming from $O_X(V)$. As such, the gluability and identity sheaf axioms are satisfied for $O_Y(U)$ because they hold for regular functions (and not because we can somehow recast them into the gluability and identity sheaf axioms for $O_x$)?

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This is straight forward. What have you tried? Please include this into your question. –  Martin Brandenburg Oct 24 '12 at 8:01