I was reading and came across this statement whose proof is said to be obvious. I however after hours still cannot figure out how to prove $S_2T(A) = S_1(S_1T(A)) = S_1T(M)$.
The definitions are:
$$(5)\quad\quad S_1T(A)= \ker T(M)\to T(P)$$ $$(6)\quad\quad S_1T(A)= \text{coker } T(Q)\to T(N)$$ and we have $$(5')\quad 0\to S_1T(A)\to T(M)\to T(P)$$ $$(6')\quad T(Q)\to T(N)\to S^1T(A)\to 0$$ and the definition of the satellites are iterated to obtain $$S_{n+1}T : = S_1(S_nT), S_0T = T, n\in\mathbb{N}$$ $$S^{n+1}T : = S^1(S^nT), S^0T = T, n\in\mathbb{N}$$
Prop 1.3 says that
If $T$ is covariant and $A$ is projective, then $S_nT(A) = 0$ for all $n>0$. If $A$ is injective, then $S^nT(A) = 0$ for all $n>0$.
Please help, I'm really stuck for a long time.
By $(5)$, $$S_{n+1}T(A)=S_1S_nT(A)=\operatorname{ker} S_nT(M)\to S_nT(P).$$ But $P$ is projective, so by Prop. 1.3, $S_nT(P)=0$, and so $\operatorname{ker} S_nT(M)\to S_nT(P)$ is just $S_nT(M)$.