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In Hatcher, the lifting criterion states (Prop 1.33):

Suppose given a covering space $p: (X^{'},x^{'}) \rightarrow (X,x_0)$ and a map $f: (Y,y_0) \rightarrow (X,x_0)$ with $Y$ path-connected and locally path-connected. Then a lift $f': (Y,y_0) \rightarrow (X^{'},x^{'})$ of f exists iff $f_*(\pi_1 (Y,y_0)) \subset p_*(\pi_1(X^{'},x^{'}))$.

Can you explain the proof of the 'only if' statement which says that this is obvious since $f_* = p_*f'_*$? How do we know that the group homomorphism $f'_*$ exists? And which property are we using to prove the existence of $f'_*$, thanks.

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It seems like they're talking about the other direction as trivial, and indeed it's fairly easy. Note since $f=p\circ f',$ functoriality gives $f_*=p_*\circ f'_*,$ and now $$f_*\pi_1(Y,y_0)=p_*f_*'\pi_1(Y,y_0)\subset p_*\pi_1 (X',x).$$

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  • $\begingroup$ So the direction that assumes $f'$ exists like @LeeMosher said. That is strange since usually I assume 'only if' to mean (<=) but in this case, it is the other way. Your answer clarified a lot thanks! $\endgroup$ – metalder9 May 21 at 17:07
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    $\begingroup$ Or am I confusing things, 'only if' usually means this direction (=>) ? $\endgroup$ – metalder9 May 21 at 17:10
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    $\begingroup$ No problem! As for your question, this might help: math.stackexchange.com/questions/1474345/… $\endgroup$ – cmk May 21 at 17:14
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The "only if" statement assumes that a lift $f'$ exists.

The definition of lift gives us the equation $f = p \circ f'$.

And the functorial properties of the fundamental group then give us the equation $f_* = p_* f'_*$.

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