# Why $f^{'}_{*}$ group homomorphism exist in 'only if' part of the lifting criterion proof?

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.

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 $$f_*\pi_1(Y,y_0)=p_*f_*'\pi_1(Y,y_0)\subset p_*\pi_1 (X',x).$$
• 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! – metalder9 May 21 at 17:07
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 $$f_* = p_* f'_*$$.