# Regular functions vs holomorphic functions

Let $$X$$ be an affine smooth variety over the complex numbers, $$X^{an}$$ its associated smooth complex analytic space, and $$\mathcal{O}$$, resp. $$\mathcal{O}^{an}$$ the respective structure sheaves.

Is the natural map

$$\mathcal{O}(X)\to \mathcal{O}^{an}(X^{an})$$

injective?

Denoting by $$f : X^{an}\to X_{Zar}$$ the morphism of sites, this map is given as the composition

$$\Gamma(X_{Zar}, \mathcal{O})\to \Gamma(X^{an},f^{-1}\mathcal{O})\to \Gamma(X^{an},\mathcal{O}^{an})$$ and in this composition we know the last map is injective (see Serre’s GAGA paper, Prop. 10(b)).

Morally, this map should send a regular function on $$X$$ to the holomorphic function it induces on $$X^{an}$$ in an evident way, so I would expect the answer to this question to be trivially yes, but I want to make sure I’m not missing anything.
If we call $$\alpha$$ the map $$\Gamma(X_{Zar}, \mathcal{O})\to \Gamma(X^{an},f^{-1}\mathcal{O})$$, then for any closed point $$x\in X$$, $$\alpha(s)_x = \alpha_x(s_x)$$, where $$s\in \Gamma(X_{Zar}, \mathcal{O})$$, $$\alpha_x$$ is the map induced on stalks, and $$x$$ is also regarded as a point of $$X^{an}$$, since analytification identifies closed points.
It is known that $$\alpha_x$$ is an isomorphism for all $$x$$ (see the discussion before Prop. 10 in loc cit), so this should be it. Am I right?
• By the strong Nullstellensatz, the zero ideal in $\mathcal{O}_X(X)$ equals the intersection of all kernels of all evaluation $\mathbb{C}$-algebra homomorphism, $$\text{ev}_x:\mathcal{O}_X(X)\to \mathbb{C}.$$ Each of these homomorphisms factors through $\mathcal{O}_{X^\text{an}}(X^{\text{an}})$. – Jason Starr 3 hours ago