Why is the Betti number for a Sphere 0, if the Euler spiral can cut a singular sphere flat? Wolfram has the Betti numbers of surfaces as follows:
**Surface Betti Number  **
cross-cap   1
cylinder    1
klein bottle    2
Möbius strip    1
plane lamina    0
projective plane    1
sphere  0
torus   2
If we were to cut a sphere with a Euler spiral, I could lay the sphere flat. Granted the cut does not meet the beginning and end points, unless we allow more of topological 'holes' rather than 'handles'(or maybe 180 flip our cut and continue back). Can some spaces not be cut as such topologically.
Are these the reasons for the Betti definition? Does it have to do with the contiguity of the cut? What is meant by a 'Noetherian Local Unit Ring'?
 A: Beyond the issues raised in my comments, the title question can be addressed by a few basic observations about topological spaces in general and the sphere in particular.
In a topological space $X$, the existence of an embedded arc (such as the Euler spiral) is irrelevant to determining the first Betti number of $X$, as you hinted in your sentence "Granted..." Instead, in a very, very rough sense the first Betti number is concerned with closed curves in $S$, meaning continuous functions $S^1 \mapsto X$ where $S^1$ is the circle.
Narrowing the focus to $X=S^2$, let's consider simple closed curves, each of which is the image of a continuous injection $S^1 \mapsto S^2$. Let $C \subset S^2$ be one such curve. According to the Jordan Curve Theorem, $S^2 - C$ is a union of two components $D_1$ and $D_2$. According to the Schönflies theorem, for $i=1,2$ the subspace $C \cup D_i$ is homeomorphic to the standard closed unit disc $\mathbb D^2$ in the plane, with $C$ correspnoding to the boundary circle of $\mathbb D^2$ and $D_i$ corresponding to the interior of $\mathbb D^2$. Clearly you can shrink that boundary circle down to a point, and this can be regarded as evidence that the first Betti number of $S^2$ is zero. Beyond that evidence, one can show that any continuous function $S^1 \mapsto S^2$ is homotopic to a constant function.
In general, in any topological space $X$, to say that $X$ is simply connected means that $X$ is path connected and every continuous function $S^1 \mapsto S^2$ is homotopic to a constant. One basic theorem of topology is that if a space $X$ is simply connected then its first Betti number is equal to zero. This is one reason why the first Betti number of $S^2$ is equal to zero.
Warning: The converse of that theorem is not true. The projective plane provides a counterexample to the converse, notwithstanding that table you got from Wolfram: the projective plane is not simply connected because it has an embedded circle which is not homotopic to a constant; but the first Betti number of the projective plane is equal to $0$.
Finally, to expand on one of my comments: the formal definition of the first Betti number of $X$ is the rank of the first homology group of $X$ with $\mathbb Z$ coefficients. From there one can, with some work, write down a formula for the first Betti number of a compact, connected surface with boundary.
