Let $(\tilde{X},p)$ be a covering space of $X$, $\tilde{x}_{0}\in\tilde{X}$, $x_{0}=p(\tilde{x}_{0})$, $y_{0}\in Y$ and $\varphi :(Y,y_{0})\rightarrow (X,x_{0})$. Under what conditions does there exist a map $\tilde{\varphi}:(Y,y_{0})\rightarrow (\tilde{X},\tilde{x}_{0})$ such that the diagram
is commutative? If we assume such a map $\tilde{\varphi}$ exists, we obtain the follow commutative diagram of groups and homomorphisms:
beacuse $p_{*}$ is a monomorphism the existence of a homomorphism $\tilde{\varphi}_{*}:\pi(Y,y_{0})\rightarrow\pi(\tilde{X},\tilde{x}_{0})$ which makes the diagram commutative, is exactly equivalent to the dondition that the image $\varphi{*}$ be contained in the image of $p_{*}$
The Massey says this before the theroem of the lifting of an arbitrary map. But I don't see where we use the fact that $p_{*}$ is a monomorphism. In general, if $$(p_{*}\circ\tilde{\varphi})=\varphi$$ the condition on the images is tha same without assumptions on $p_{*}$... If I know that there is that commutative diagram, my necessary condition is image of $\varphi_{*}$ in image of $p_{*}$. I don't need the fact that $p_{*}$ is a monomorphism. Is it wrong?
Given $\varphi_*$ and $p_*$, there may be no $\tilde{\varphi}_*$ such that $(p_* \circ \tilde{\varphi}_*) = \varphi_*$. It is the existence of such a $\tilde{\varphi}_*$ that is dependent on the condition that the image of $\varphi_*$ is contained in the image of $p_*$.