## The Fundamental Groupoid and Group (Pt. II)

**Point of Post: **This is a continuation of this post.

We first verify that this pairing satisfies some of the properties we expect out of it:

**Theorem: ***Let be a topological space. If , , and denote the constant paths and , then . *

**Proof: **To prove that define by

It is clear that is continuous by the gluing lemma, since the overlap of the domains is where the second case gives $y$ and the first gives

Moreover,

and

Thus, we may conclude that and thus .

The other case follows similarly.

Intuitively, we can think of this homotopy as doing for less, and less time until we are eventually only doing it for the instant .

For each path let denote the path defined by . This is the path obtained by doing in the reverse. It should seem fairly obvious that if we start at a point, make a path to another point, and then take the same path in reverse, that (up to homotopy) we just get the initial path at the point. More formally:

**Theorem: ***Let . Then, for any one has that and .*

**Proof: **Once again, we prove only that . Indeed, define by

Once again, one can verify that all the boundary conditions are in agreement, and so is continuous by the gluing lemma. Moreover, one can verify that, indeed, and that . Thus, and so .

Last, but certainly not the least annoying, is the verification that (like all good products should be) the multiplication is associative. Indeed:

**Theorem: ***Let and . Then, *

**Proof: **As per usual, we define a homotopy . In particular, define according to the rule

I, once again, leave the busy work to the reader to check that the conditions agree on overlap, and thus is continuous. Moreover, I leave it to you lucky folks to check that, in fact, and .

Now that we have verified that our product satisfies such nice properties, we can, in fact, create a fairly tame structure. In particular, call a category a *groupoid *if every arrow is an isomorphism. We claim then that if , called the *fundamental groupoid of *, is the category whose object set , for which , and for which composition is taken by the product then:

**Theorem: *** is a groupoid.*

**Proof: **It’s evident that is a category, whose identity at each object is merely . We see then that each arrow is indeed invertible since is an inverse.

**References:**

[1] Spanier, Edwin Henry. *Algebraic Topology*. New York: McGraw-Hill, 1966. Print.

[2] Hatcher, Allen. *Algebraic Topology*. Cambridge: Cambridge UP, 2002. Print.

[3] Bredon, Glen E. *Topology and Geometry*. New York: Springer-Verlag, 1993. Print.

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