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Show up a t q ͐ t s s ̃j v b v ł b ̏c , h , e p c ̍w a t ܂Łb ς ͖ Ŏt Ă ܂ b fujitsubo akakimoto aapexera ablits amugen a hks atrust aganador aauto exe asignal. 1(X) finite, then every map X → S1 is nullhomotopic Use the covering space R → S1 Solution Call the map in question f π 1(X) is finite, so the image of f ∗ is finite The only finite subgroup of π 1(S1) ≈ Z is 0 So the lifting criterion, Proposition 133, tells us that there’s a lift fe X → R to the universal cover R. Archibald craven mrs sowerby martha ben weatherstaff bedridden orphan misselthwaite dr craven colin maidservant hunchback india mrs medlock health key.
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