If we have the map $\phi_s$ that takes $\overline{1}\in\mathbb{Z}/p^a$ to $\overline{p^{s+b}/d}\in\mathbb{Z}/p^b$, where $d=\gcd(p^a,p^b)$. Why in order to see that $\phi_s$ is a well-defined map, we need to show that $\overline{p^a}\phi_s(\overline{1})=\overline{0}\in\mathbb{Z}/p^b$?
EDIT:
Is the meaning of well defined if we have $\mathbb Z_{n}'s$ as domain and target is just to send the identity to the identity?If so, for me, that definition of well defined is very different from the one ordinary used (if $x=y$ then $f(x) = f(y)$)
Is there any connection between both definitions?
It seems that the question implicitly requires the map to be an abelian group homomorphism, i.e. $\phi_s(x+y)=\phi_s(x)+\phi_s(y)$ for all $x,y$ in the domain. In particular, if we write $n\cdot x$ for $x+\ldots+x(n$ times$)$ (which is essentially viewing abelian groups as $\def\Z{\mathbb Z}\Z$-modules), then $\phi_s(n\cdot x)=n\cdot\phi_s(x)$. In this case it is the same as $\bar n\phi_s(x)$.
For the map to be well-defined, the problem is $\phi_s(\bar0)$. In $\Z/p^a\Z$, $\bar0=\bar{p^a}=p^a\cdot\bar1$. Hence $\bar0=\phi_s(\bar0)=\phi_s(p^a\cdot\bar1)=p^a\cdot\phi_s(\bar 1)=\bar p^a\phi_s(\bar 1)$, which is what you want to check.