This is a question on Cartan Eilenberg Homological Algebra Pg 42, Satellite.
Suppose given the following exact sequences of modules with $P$ projective over some $\Lambda$ ring.
$0\to A'\to R\to P\to 0$
$0\to A'\to A\to A''\to 0$
Further suppose $0\to M\to P\to A\to 0$ and $0\to M\to R\to A\to 0$ are exact s.t. the arrows induced $P\to A''$ and $R\to A$ form commutative diagram. This induces a unique arrow on $A'\to A$. Assume $M\to M$ arrow is identity arrow and $M\to M,M\to P,R\to P$ and $M\to R$ forms a commutative square.
Let $T$ be a half exact additive functor.(i.e. Given $0\to A\to B\to C\to 0$, then $TA\to TB\to TC$ is exact.)
Then $T$ acts on the diagram formed by previous 4 exact sequence yields the following exact sequences.
$0\to T(M)\to T(M)\to 0$
$0\to T(A')\to T(R)\to T(P)\to 0$ is exact by $P$ projective and $T$ additive.($0$ of first exact sequence is aligned with $T(A')$ position.)
There will be arrows $0\to T(A'),T(M)\to T(R)$ and $T(M)\to T(P)$. Then Snake lemma yields $Ker(T(M)\to T(P))\to T(A')\to Coker(T(M)\to T(R))$ is exact.
$\textbf{Q:}$ The book says "$T(M)\to T(R)\to T(A)$ is exact, it follows $Ker(T(A')\to Coker(T(M)\to T(R)))=Ker(T(A')\to T(A))$." How is this supposed to be obvious? I did verify the equality but I do not see any particular reason this is obvious. Or was kernel universal property invoked here?
I think the following is the most direct reasoning, although perhaps you already reasoned this way:
As $T(M)\xrightarrow{f} T(R)\xrightarrow{g} T(A)$ is exact, $\operatorname{coker}(f)\rightarrow T(A)$ is monic; thus for any $X\rightarrow T(R)$, we have $\ker(X\rightarrow \operatorname{coker(f)}) = \ker(X\rightarrow T(A))$.