fs-an-unbounded-total-derived-functor-exists-from-enough-injectives-alone.md

Assume AC and put $R=\mathbb{Z}\mathrm{/}4$. Its ideals are $0,2R,R$. An $R$-map $2R\to R$ sends $2$ to $0$ or $2$, so it extends by multiplication by $0$ or $1$. Maps on $0$ and $R$ extend trivially. Baer's criterion therefore makes $R$ injective. The doubly infinite complex $I^i=R,d^i=2$ is termwise injective and acyclic since kernel and image of two both equal $2R$.

For $F={\mathrm{Hom}}_R(R\mathrm{/}2,-)$, each $F(I^i)$ is $2R\cong \mathbb{Z}\mathrm{/}2$, and its differential is zero. Thus $F(I)$ is not acyclic. The quasi-isomorphism $0\to I$ is a termwise-injective replacement of the zero complex, but this recipe sends it to a nonzero derived object, while replacement by zero gives zero. The unbounded recipe is therefore not well defined.

Indeed $I$ is not K-injective: if its identity were nullhomotopic, the degreewise equation would be $1=2a_i+2a_{i+1}$, impossible modulo two. An acyclic complex must have zero Hom into a K-injective target, so taking the source to be $I$ violates that condition. The bounded-below construction avoids this example through its boundedness hypothesis. This refutes the arbitrary-replacement assertion, not existence of unbounded derived functors by other methods.