# If every holomorphic function $f$ defined on an open subset $\Omega$ of $\mathbb C$ has a primitive. Is it true that $\Omega$ is simply connected?

If every holomorphic function $$f$$ defined on an open subset $$\Omega$$ of $$\mathbb C$$ has a primitive. Is it true that $$\Omega$$ is simply connected?

I saw this post: If every harmonic function on $$\Omega$$ has a harmonic conjugate on $$\Omega$$, then $$\Omega$$ is simply connected. Then I am curious whether it is true for assuming the existence of primitive.

I think we can decompose an open set into the union of connected components. Then suppose there is a connected component that is not simply connected. By the definition of simply connected, there exists a point $$z_0\notin\Omega$$ encompassed by a Jordan curve $$\gamma$$ lying in a small circle which is completely contained in the connected component. Consider the holomorphic function $$f=\frac 1{z-z_0}$$, it should be clear that $$f$$ does not have a primitive in the small neighborhood. Contradiction!

Is my proof correct?

Edit: I should have claimed that every connected component of $$\Omega$$ is simply connected.

Assuming connectedness, you are making things too complicated. Rudin's RCA has a theorem giving various ways of characterizing simply connected regions and this is one of them. If you know that a region is simply connected iff the integral of any analytic function over any closed path in it is $$0$$ then this result is obvious because the integral of a derivative over a close path is $$0$$ by the very definition of the integral.