Is this a morphism of sheaves? Consider the complex plane $\mathbb{C}$, equipped with the sheaf $\mathcal{F}$ of holomorphic functions  $\mathbb{C} \to \mathbb{C}$. I'm looking at the following map of sheaves
$$ \varphi: \mathcal{F} \to  \mathcal{F}, \, \, \, f \mapsto f^2 + 1.$$
I am trying to figure out if this actually is a map of sheaves. (That is, whether the map $\varphi$ commutes with the various restriction maps on $\mathbb{C}$.) My gut tells me that $\varphi$ should be a map of sheaves; I mean, $\varphi$ is a polynomial in it's input! But when I go to check that $\varphi$ commutes with the restriction maps, it seems impossibly complicated. It would involve knowing exactly how $\varphi$ acts every open set of the complex plane. So far, I've tried showing that $\varphi$ commutes with the restriction map from $\mathbb{C}$ to the open unit disk. This already is quite complicated, since it involves figuring out how $\varphi$ acts on $D$, which isn't obvious at first sight. I have two questions here:

*

*Is there an easy way to check that this map commutes with the restriction maps? I'm looking for something along the lines of: "if $\varphi$ is a polynomial in its input, then it commutes with restriction maps." Or is there no general rule like this, in which case we have to check each example by hand?


*More generally, is there an easy test (like the one I mentioned above) that we can use to generate many examples of maps of sheaves? This could be a test that applies for this particular example of a sheaf, or other sheaves as well (e.g: continuous functions on $\mathbb{R}^n$, smooth functions on a manifold, etc.)
Thanks for the help.
 A: The direct method is actually quite simple – you may be lost in the formalism though.
I’m of course assuming that you’re assuming that $\mathcal{F}(U)$ is, for each open subset $U$, the space of holomorphic functions $U \rightarrow \mathbb{C}$ and that the restriction of $\mathcal{F}$ is the usual notion of restriction for a function. (You didn’t specify otherwise, even though it is important.)
If $U \subset V \subset \mathbb{C}$ are nonempty open subsets, and if $f \in \mathcal{F}(V)$, then $\varphi(V)(f)=f^2+1$, so that its restriction to $U$ is $h:z \in U \longmapsto f(z)^2+1$. On the other hand, the restriction $g$ of $f$ to $U$ is $z \in U \longmapsto g(z)=f(z)$. Now $\varphi(U)(g)$ is $z \longmapsto (g^2+1)(z)=g(z)^2+1$, so that $\varphi(U)(g)=h$, QED.
A: Let $U \subseteq V$ be open subsets of the complex plane, so that we have a restriction mapping $f \mapsto f|_{U} : \mathcal F(V) \to \mathcal F(U)$. Note that the definition of $f|_U$ is given by $f|_U(z) = f(z)$. Now let $f \in \mathcal F(V)$ be arbitrary. We have to check that $(f|_{U})^2 + 1 = (f^2 + 1)|_{U}$. To see that two functions on $U$ agree, let us take $z \in U$ arbitrary. We then have
$$
((f|_U)^2 + 1)(z) = (f|_U)^2(z) + 1 = (f|_U(z))^2 + 1 = f(z)^2 + 1 = (f^2 + 1)(z) = (f^2 + 1)|_U(z).
$$
It is worth emphasizing that essentially no mathematics happens in this sequence of equations: it is just seeing through all the notation.
