Well, so far, I have noticed that whenever a matrix lie group is connected it is path connected, so is it true that in matrix lie group connected $\Rightarrow$ path connected?If yes, could anyone tell me where I can get the proof?or if some one tell me the sketch of the proof. Thank you.
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This is true for any locally path-connected space (this is the crucial property to use in the proof), in particular any manifold. Slightly more generally, for locally path-connected spaces, components and path components coincide.
Qiaochu Yuan
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Like Qiaochu Yuan said, any connected locally path-connected space is path-connected. This is because local path-connectedness implies that the path-connected components are open (this is essentially by definition: every point admits a path-connected neighbourhood, and hence is an interior point of its path-connected component), and therefore are also closed (since the complement of a component is the union of the other component, hence a union of open sets). Therefore, by connectedness, there is only one path-connected component and it is everything.
Cronus
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Of course you are right, can't imagine what I'd been thinking. I corrected the mistake. – Cronus Jun 12 '19 at 07:20
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Nice answer!!${}$ – cqfd Jun 12 '19 at 07:30
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1Thanks! I am glad it wasn't completely pointless to answer this six year old question ;) – Cronus Jun 12 '19 at 07:32