Germ at any point determines analytic function

If I remember by Riemann surfaces course correctly, then the following should be true:

Let $X$ be a Riemann surface, $U\subset X$ be open, and $x\in U$. Then the map $\Gamma(U,\mathcal{O}_X)\to \mathcal{O}_{X,x}$ is injective.

This should follow from the uniqueness of analytic continuation. Of course the map is not surjective, as can be seen by considering $X=\mathbb{C}^*$, $x=1$, and taking the germ of a logarithm.

My question is if this theorem is still true on higher dimensional complex manifolds? So:

Let $X$ be a complex manifold, $U\subset X$ be open, and $x\in U$. Then the map $\Gamma(U,\mathcal{O}_X)\to \mathcal{O}_{X,x}$ is injective.

Also, a related question:

Let $X$ be a complex manifold, $U\subset X$ be open and simply connected, and $x\in U$. Then the map $\Gamma(U,\mathcal{O}_X)\to \mathcal{O}_{X,x}$ is an isomorphism.

Are these statements true?

• Yes, to the first question. Analytic continuation is unique just as in one dimension. Your final question: that fails even in one complex variable. – Lord Shark the Unknown Aug 12 '18 at 10:23
• @LordSharktheUnknown Thanks for the answer. Yes I realise now that the last question was kind of silly. – user2520938 Aug 12 '18 at 10:25
• @LordSharktheUnknown Be careful with what you mean by analytic continuation. The notion is highly non-trivial. Are you permitting multi-valued extensions? – AmorFati Oct 26 '18 at 6:43