Lecture notes for Friday, Feb. 24

van Kampen’s Theorem for the fundamental groupoid. Let \(X\) be a space and \(U\) and \(V\) two open subsets of \(X\) such that \(X = U \cup V\). Then the following is a pushout square of groupoids:

\(\pi_{\le 1}(U \cap V)\)	\(\to\)	\(\pi_{\le 1}(U)\)
\(\downarrow\)		\(\downarrow\)
\(\pi_{\le 1}(V)\)	\(\to\)	\(\pi_{\le 1}(X)\)

There is also a version for the fundamental groupoid on a subset of the basepoints, \(\pi_{\le 1}(X,A)\), but \(A\) needs to satisfy the following condition: \(A\) contains at least one point in each component of each of \(U \cap V\), \(U\) and \(V\). It is actually this version that we are really after, since the whole fundamental groupoid is impractical in that it contains a lot of redundant information.

Our strategy for proving this version will be to deduce it from the version for the full fundamental groupoid. The hypothesis on \(A\) guarantees that for \(W = U\cap V, U, V, X\) the groupoid \(\pi_{\le 1}(W,A)\) is equivalent to \(\pi_{\le 1}(W)\) and one might hope that replacing each groupoid in a pushout square by an equivalent groupoid yields a new pushout square. This is not quite right: replacing each groupoid by an isomorphic groupoid would of course give a new pushout square, but isomorphisms are precisely the relation between objects that pushouts are meant to preserve, they won’t preserve mere equivalence. There is a more flexible notion of pushout for groupoids (variously called, weak pushout, homotopy pushout or 2-categorical pushout) that is invariant under equivalence, but we won’t talk about that in this tutorial.

Instead what we’ll do to deduce the version for \(\pi_{\le 1}(\_,A)\) from the version for the full groupoid is (1) show that the commutative square

\(\pi_{\le 1}(U \cap V,A)\)	\(\to\)	\(\pi_{\le 1}(U,A)\)
\(\downarrow\)		\(\downarrow\)
\(\pi_{\le 1}(V,A)\)	\(\to\)	\(\pi_{\le 1}(X,A)\)

is a retract of the square for the full fundamental groupoids and (2) using the general category theoretical fact that a retract of a pushout square is also a pushout square.

A retract of an object \(X\) is another object \(Y\) with a pair of morphisms \(i : Y \to X\) and \(r : X \to Y\) such that \(r \circ i = \id\). We think of \(i\) as including \(Y\) inside \(X\) and \(r\) as being a retraction of \(X\) onto \(Y\). When we say a commutative square \(\rho\) is a retract of another square \(\sigma\) we mean that each corner of \(\rho\) is a rectract of the corresponding corner of \(\sigma\), but more than just this: all of the inclusions & retractions have to be compatible with one another in the sense that the cubical diagram formed by the two squares and the four inclusions commutes, as does the cube formed by the two squares and the four retractions. It only takes, as one says, a straight-forward diagram chase to prove that a retract of a pushout square must also be a pushout square.

Now, in our case it is easy to show the second square is a retract of the first. The inclusions are just that: inclusions \(\pi_{\le 1}(X,A) \to \pi_{\le 1}(X)\). The retractions are built as follows. To retract \(\pi_{\le 1}(X)\) onto \(\pi_{\le 1}(X,A)\) just pick, for every point \(x \in X\) a path \(\alpha_x\) from \(x\) to some point \(a \in A\), but do this in such a way that if \(x\) is already in \(A\), \(\alpha_x\) is the identity morphism at \(x\). (We can always pick these paths because the hypothesis include that \(A\) has at least one point in each component of each of the spaces we use.) Then the retraction we’re defining is the morphism that sends each \(x\) to the other endpoint of \(\alpha_x\), and each morphism \(\beta : x \to y\) to the morphism \(\alpha_y \circ \beta \circ \alpha_x^{-1}\). To ensure that the cube formed by the two van Kampen squares and the four retractions commutes, simply always pick the same \(\alpha_x\) for \(x\) in all of the groupoids it appears in.