I'm reading Gelfand's Methods of homological and here's a problem:
Moreover, as the functor $X \mapsto(X, 0) $ represents $ \mathcal{A} $ as a full subcategory of $ \mathcal{A}[T]$, $X $ is projective in $ \mathcal{A} $ for any projective $(X, t) $ in $ \mathcal{A}[T]$.
Here $\mathcal{A}$ is an abelian category and $\mathcal{A}[T]$ is defined as follow:
$\operatorname{Ob} \mathcal{A}[T]=\left\{\operatorname{pairs}(X, t) \text { with } X \in \operatorname{Ob} \mathcal{A}, t \in \operatorname{Hom}_{\mathcal{A}}(X, X)\right\}$
A morphism $(X, t) \rightarrow\left(X^{\prime}, t^{\prime}\right)$ in $\mathcal{A}[T]$ is a morphism $f : X \rightarrow X^{\prime}$ in $\mathcal{A}$ such that $t^{\prime} \circ f=f \circ t$.
I can't prove this claim. Given an exact sequence $Y \stackrel{g} \rightarrow X^{\prime} \rightarrow 0$ and $X \stackrel{f} \rightarrow X^{\prime}$, I cannot construct a morphism $(X,t) \stackrel{f} \rightarrow (X^{\prime},t^{\prime})$, so I cannot use the projectivity of $(X,t)$ to prove the projectivity of $X$.
Any hint please? Thank you.