Let $M,N,P$ be $R$-modules ($R$ commutative ring with $1$) and let $f:M\to N$ be a $R$-module homormorphism. Let tensor the homomorphism to get $ f \otimes 1_P : M \otimes P \to N \otimes P $.
I want to know the relation between $\operatorname{Coker}(f)$ and $\operatorname{Coker}(f \otimes 1_P)$.
A basic fact in homological algebra is that the functor $$-\otimes_R P:\mathsf{Mod}_R\to\mathsf{Mod}_R$$ is always right exact. So, applying $-\otimes_RP$ to an exact sequence $$ M\xrightarrow{f}N\to Q\to 0\tag{1} $$ gives an exact sequence $$ M\otimes_R P\xrightarrow{f\otimes_R\DeclareMathOperator{id}{id}\id_P}N\otimes_RP\to Q\otimes_R P\to 0\tag{2} $$ Now, $Q=\DeclareMathOperator{Cok}{Cok}\Cok(f)$ always fits in (1) so (2) gives an isomorphism $$ \Cok(f\otimes_R \id_P)\simeq\Cok(f)\otimes_R P $$
There is a short exact sequence $M\xrightarrow{f} N\xrightarrow{g}\operatorname{coker} f\to 0$. Applying the functor $(\mathord-)\otimes_RP$, which is right exact, we get a new exact sequence $M\otimes_RP\xrightarrow{f\otimes1_P} N\otimes_RP\xrightarrow{g\otimes1_P}(\operatorname{coker} f)\otimes_RP\to 0$
