By Luigi A. Rosati

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For any two elements of Idem(C(X)), choose representatives E in M(m, C(X)) and F in M(n, C(X)) and deﬁne [E] + [F] = [diag(E, F)]. Then Idem(C(X)) is an abelian monoid. Proof Because diag(SES−1 , TFT−1 ) = diag(S, T) diag(E, F) diag(S, T)−1 for all S in GL(m, C(X)) and T in GL(n, C(X)), we see that addition respects similarity classes. 1 gives us [E] + [F] = [diag(E, 0)] + [diag(F, 0)] = [diag(E, 0, F, 0)] = [diag(E, F, 0, 0)] = [diag(E, F)], and therefore addition on Idem(C(X)) is well deﬁned.

Set Ox = fx−1 (GL(n, C)). 6 we know the set GL(n, C) is open in M(n, C), so Ox is open in Ux . Moreover, because Ux is open in X, the set Ox is open in X and the union of the sets Ox is an open subset of X. To complete the proof, note that for every x in X and y in Ux the 28 Preliminaries matrix fx (y) is invertible if and only if γy is an isomorphism. Thus Ox , O= x∈X and therefore O is open in X. 21 Let X be a compact Hausdorﬀ space and suppose A1 and A2 are closed subspaces of X whose union is X and whose intersection Z is nonempty.

3 Deﬁne φ1 : Z2 −→ Z4 as φ1 (0) = 0 and φ1 (1) = 2, and deﬁne φ2 to be the homomorphism that maps 0 and 2 to 0, and maps 1 and 3 to 1. Then the sequence 0 / Z2 φ1 / Z4 φ2 / Z2 /0 / G3 /0 is exact. 4 A short exact sequence 0 / G1 φ1 / G2 φ2 40 Preliminaries is split exact if there exists a group homomorphism ψ : G3 −→ G2 such that φ2 ψ is the identity map on G3 ; we call such a homomorphism ψ a splitting map. We write split exact sequences using the notation 0 / G1 φ1 / G2 g / G3 φ2 / 0. 2 is split exact because the homomorphism ψ : H −→ G ⊕ H given by the formula ψ(h) = (0, h) is a splitting map.