Questions about representation theory of associative algebras. I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1. 
I have two questions on page 85. On Line 18 of Page 85, it is said that $\ker p_i \subseteq \operatorname{rad} P_i$. I know that if $\ker p_i \subseteq \operatorname{rad} P_i$, then $\ker p_i$ is superfluous. Since the sequence in the proof is a minimal projective resolution of $S$, $\ker p_i$ is superfluous. How to show that if $\ker p_i$ is superfluous, then we have $\ker p_i \subseteq \operatorname{rad} P_i$?
On Line -6 of Page 85, it is said that $$ \hom _A(\operatorname{rad} P(a)/ \operatorname{rad}^2 P(a), S(b)) \cong  \hom _A(\operatorname{rad} P(a)/ \operatorname{rad}^2 P(a), I(b)). \quad (1) $$ It seems that we do not have $S(b) \cong I(b)$, how to show that (1) is true? Thank you very much. I attached Page 85 of the book. 
 A: (1) The first statement can be proven by contradiction. The important fact for your first question is that the indecomposable projectives have a simple top. Thus, let $P_i:=\oplus_j P_i^j$, where $P_i^j$ is an indecomposable module. Suppose $\operatorname{ker}p_i\nsubseteq \operatorname{rad} P_i$. Then, since $\operatorname{rad} P_i^j$ is the unique maximal submodule of $P_i^j$ there exists at least one $j$ such that $P_i^j\subseteq \operatorname{ker} p_i$. But now $\ker p_i+\oplus_{k\neq j} P_i^k=P_i$ and the second summand is a proper submodule. Hence $\ker p_i$ is not superfluous.
(2) Here the important fact is that $\operatorname{rad} P(a)/\operatorname{rad}^2 P(a)$ is a semisimple module. Because of the fact that $\operatorname{Hom}(M,-)$ is left exact it holds that $\operatorname{Hom}(M,S(b))\hookrightarrow  \operatorname{Hom}(M,I(b))$ for any module $M$. But now since $\operatorname{rad} P(a)/\operatorname{rad}^2 P(a)$ is semisimple we know that $\operatorname{soc} ( \operatorname{rad} P(a)/\operatorname{rad}^2 P(a) )=\operatorname{rad} P(a)/\operatorname{rad}^2 P(a)$. And we know that $f(\operatorname{soc} M)\subseteq \operatorname{soc} N$ for any morphism $M\to N$. In this special case we can deduce $f(\operatorname{rad} P(a)/\operatorname{rad}^2 P(a))\subseteq \operatorname{soc} I(b)=S(b)$ and thus the map is also surjective.
