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Starting with fundamental group, say of a circle, let's reflect back to path groupoid a little. The path concatenation operation is partial, but this can be remedied by focusing on the sets of paths, rather than individual paths. Then, if the two sets of paths of the multiplier have no end points coinciding with start points of the multiplicand, then the result is the empty set, which is the additive identity of the ring. Therefore, Boolean algebra of sets provides the additive structure of the ring.

The ring multiplication is inherited from groupoid, that is path concatenation, and it is non commutative. There is a multiplicative identity, which is the set of trivial loops around all points in the space.

To prove the distributivity property, let $P$,$Q$,$R$ be sets of paths. Then

$P (Q \cup R) = \{p \cdot x | p \in P \wedge x \in Q \cup R \} = $

$= \{p \cdot x | (p \in P \wedge x \in Q) \vee (p \in P \wedge x \in R) \} = $

$= \{p \cdot x | (p \in P \wedge x \in Q) \} \cup \{p \cdot x | (p \in P \wedge x \in R) \} $

At this point, it is not immediately obvious if any congruence is needed to be introduced. Yet, I'm puzzled why there appears nothing on the web matching the "fundamental ring" search term. Or there is? And what is the ring of path sets of the circle, is it a familiar mathematical object?

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    $\begingroup$ If I understand your definition correctly (and I'm absolutely not sure whether I do, it would help if you gave a clear definition of what you mean by fundamental ring), then it doesn't sound like the fundamental ring will be a homotopy invariant. Compare for example the interval and the point. The fundamental groupoids of two homotopy equivalent spaces are equivalent (in the sense of equivalence of categories) $\endgroup$
    – Achim Krause
    Commented Jul 24, 2020 at 17:22
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    $\begingroup$ Why does multiplication distribute over addition? $\endgroup$
    – LSpice
    Commented Jul 24, 2020 at 17:47
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    $\begingroup$ If you are using union for addition you are looking at a semiring or quantale structure. The subsets of a groupoid form an involutive quantale. See the work of Pedro Resende $\endgroup$
    – Benjamin Steinberg
    Commented Jul 24, 2020 at 23:38
  • $\begingroup$ This seems like it could be the groupoid ring of the fundamental groupoid with $\mathbb Z/2\mathbb Z$-coefficients, but it is entirely unclear from the phrasing of the question. $\endgroup$
    – Joshua Mundinger
    Commented Jul 25, 2020 at 2:21
  • $\begingroup$ @Benjamin You are right, the fact that it is semiring, rather than ring dampens my enthusiasm considerably. (I'll keep incorrect title because of the pun in the title). $\endgroup$
    – Tegiri Nenashi
    Commented Jul 25, 2020 at 22:59

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