Suppose that $(X,\mathcal{O}_{X})$ is a variety and $Y\subset X$ is closed and equipped with the induced subspace topology. Consider the variety $(Y,\mathcal{O}_{Y})$, where for every open $V\subset Y$ we have $\mathcal{O}_{Y}(V):= \{f:V\rightarrow k\rvert \forall P\in V, \ \exists U\subset X, \ \exists g\in\mathcal{O}_{X}(V) \ \text{s.t} \ P\in U, \forall Q\in V\cap U \ \text{we have} \ f(Q) = g(Q)\}$. Let $i:Y\rightarrow X$ be the inclusion map, and notice that it is a morphism. Let $(Z,\mathcal{O}_{Z})$ be a variety and $f:Z\rightarrow Y$ a map of sets. I want to prove that $f$ is a morphism if and only if $i\circ f$ is a morphism.
My Attempt: "$\implies$" Suppose that $f$ is a morphism, then it follows immediately from the fact that $k$-spaces and their morphism form a category that the composition of two morphisms is again a morphism, so $i\circ f$ is a morphism.
"$\impliedby"$ Suppose that $i\circ f$ is a morphism. One can easily proof that $f$ is continuous. To prove that it is also a morphism we have to prove that for every $V\subset Y$ open and $\varphi\in\mathcal{O}_{Y}(V)$ we have $\varphi\circ f\in\mathcal{O}_{Z}(f^{-1}(V))$. If $V\subset Y$ open, then $\exists U\subset X$ open such that $V=U\cap Y = i^{-1}(U)$. Thus if $\varphi\in\mathcal{O}_{Y}(V)$, then $\varphi\in\mathcal{O}_{Y}(i^{-1}(U))$. My claim would be that I can somehow extend $\varphi$ to a regular function $\psi$ in $\mathcal{O}_{X}(U)$, since then I can use the fact that $i\circ f$ is a morphism, which would imply that $\psi\circ i\circ f\in\mathcal{O}_{Z}(f^{-1}(V))$, and then one can use the fact that $\psi\circ i = \varphi$. But I have problems proving that the extension of $\varphi$ to a regular function in $\mathcal{O}_{X}(U)$ is possible.
Extra comment: If $\varphi\in\mathcal{O}_{Y}(V)$, then let $P\in V=i^{-1}(U)$, then there exists $U_{P}\subset X$ open and $g\in\mathcal{O}_{X}(U_{P})$ such that $P\in U_{P}$ and $\forall Q\in U_{P}\cap V = U_{P}\cap i^{-1}(U)$ we have $\varphi(Q) = g(Q)$. But this doesn't imply that $g\in\mathcal{O}_{X}(U)$ and also not that for all $Q\in i^{-1}(U)$ we have $\varphi(Q) = g(Q)$.
The missing ingredient here is that to check $\varphi \circ f \in \mathcal{O}_Y(f^{-1}(V))$ it is sufficient to know the same fact for a sufficiently small neighborhood of any point in $V$. That follows immediatly from your extra comment.