Look at the proof of Morera. All you need is to establish path-independence, so any family of closed curves will do. Now tell me why $\gamma$ doesn't go around the origin, so your complaint is not valid :) [Take your $C$ to be a triangle if you wish.]
I mistyped, though; you're right. $f(w)/w^2$ likely does have a pole at the origin. But it's not relevant.
@Thorgott But then how do you know that we have a topological push-out? Did you explicitly compute the prime ideals? I actually don't know how the primes generally look like for a pull-back $A\times_C B$ or rings, except when $C=0$.
@Koro They'll get the points when the suspension finishes.
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@ShaVuklia Hmm, I see. The issue is that the inclusion of affine schemes into schemes does not preserve pushouts.
Whatever, let's scratch the dumb algebraic geometry.
Take $X$ to be the $3$-point space with one open and two closed points. A sheaf on this space is the same data as a pullback square (the global sections are the pullback of the maps from the sections on the two open subsets with two points to the sections over their intersection, which is the open point). Take the sheaf corresponding to the pullback of the two projections $\mathbb{Z}\rightarrow\mathbb{Z}/6\mathbb{Z}$. Take the section $3$ in one copy, the section $0$ in the other copy and the global section $(2,2)$. Restricting the global section to each copy of $\mathbb{Z}$ and multiplying…
i assumed that any element $U_i$ of a cover $\mathcal U$ containing $x \in X$ contains at least one neighborhood of $x$ that is a subset of $X$, hopefully i break nothing
a more readable statement: suppose $X$ is a metric space and $x \in X$. Any element $U_i$ of an open cover $\mathcal U$ of $X$ that has $x \in U_i$ also has as a subset at least one neighborhood $B \subseteq X$ of $x$
this is absolutely uncontroversial, nothing will explode
