By Smirnov V. A.

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**Extra info for A-Simplicial Objects and A-Topological Groups**

**Sample text**

We observe that all the groupoid products are defined, and that ϕn = cn−1 (1) · ϕn−1 for any n ∈ Z. 1 The map c : R → G is continuous and satisfies the conditions that c (0) = 1c(0) , and that αc (t) = c(0) and βc (t) = c(t) for all t ∈ R. Proof It is immediate that c is continuous at all points except, possibly, integer points n ∈ R. It is also immediate that limt→n+ c (t) = c (n); to demonstrate continuity from the left we note that lim c (t) = lim cn−1 (t − (n − 1)) · ϕn−1 = cn−1 (1) · ϕn−1 t→n− t→n− using the continuity of cn−1 on I, whereas c (n) = cn (0) · ϕn = 1cn (0) · ϕn = ϕn = cn−1 (1) · ϕn−1 .

It follows that there are only finitely many values of t for which 34 2 Connections on Lie Groupoids and Lie Algebroids c is not smooth. At the points where c is not smooth, all left and right derivatives must nevertheless exist because they do for the curves cn and cn ; thus c is piecewise smooth. It is clear that a similar approach can be used to give the lift of a curve defined on an arbitrary open interval containing zero. We may then define the lift of a vector field by lifting the curves in its flow.

1 we may assume that each c [0, 1] ∩ Ui is connected and that the Ui are indexed so that c(0) ∈ U1 and c(1) ∈ Um , and so that we may find c(ti ) ∈ Ui ∩ Ui+1 for 1 ≤ i ≤ m − 1 and 0 < t1 < · · · < tm−1 < 1; put t0 = 0 and tm = 1. 2 Infinitesimal Connections on Lie Algebroids 39 where the image of each subinterval lies completely within one of the trivialization neighbourhoods U. We shall demonstrate the existence of a unique lifted curve segment c satisfying the condition γ c˙ (0) = c˙ (0) and defined on the first of these intervals [0, a] where a ≤ t1 ; this will be the first segment of a lifted curve defined on the whole of [0, 1].