If $\Gamma$ is $\Lambda$-projective and $C$ is $\Gamma$-injective, then $C$ is $\Lambda$-injective.

Question

This is a problem I ran into while reading Cartan Eilenberg's Homological algebra pg 30.

Given a unital ring homomorphism $\varphi:\Lambda\to\Gamma$, I want to prove the underlined statement.

I tried to mimic the proof of 6.2, but to no avail. For convenience I cite the proof here.

My thoughts:

Because of natural isomorphism

$$\hom_\Lambda(A,C) = \hom_\Gamma(A_{(\varphi)},C),$$ where $A_{(\varphi)} = A\otimes_\Lambda \Gamma$, we only need to prove that $A_{(\varphi)}$ is an exact functor in $A$. It follows that $\hom_\Gamma(-_{(\varphi)},C) = \hom_\Lambda(-,C)$ is exact, therefore $C$ is $\Lambda$-injective. We already know that $A_{(\varphi)}$ is right exact, all that is left is to show that $i:A'\to A$ injective implies that $$i\otimes \Gamma: A'\otimes_\Lambda \Gamma\to A\otimes_\Lambda \Gamma$$ is injective using the fact that $\Gamma$ is $\Lambda$-projective.

Answer 1

Maybe it's helpful to explain the general principle here, because it is likely that you will encounter statements like this in many other contexts as well:

If a functor $U:{\mathscr A}\to{\mathscr B}$ between abelian categories has an exact left adjoint, then it preserves injectives. This applies to the forgetful functor $U: \Gamma\text{-Mod}\to\Lambda\text{-Mod}$ here, since its left adjoint $(-)_{(\varphi)} := \Gamma\otimes_\Lambda -$ is exact if $\Gamma$ is $\Lambda$-projective (and in fact, flatness would suffice, as user26857 said).

The dual of this statement (which is now really a formal consequence and need not be reproven) is: If $U:{\mathscr A}\to{\mathscr B}$ has an exact right adjoint, then is preserves projectives. This applies to $U$ and its right adjoint $\text{Hom}_\Lambda(\Gamma,-)$, which is exact if and only if $\Gamma$ is $\Lambda$-projective.

Also, note that relative projectivity and injectivity can be formalized in this generality: given $U: {\mathscr A}\to{\mathscr B}$ exact, call $X\in{\mathscr A}$ $U$-injective or $U$-projective if it is injective resp. projective relative to all monomorphisms resp. epimorphisms in ${\mathscr A}$ that split upon application of $U$. Then any ${\mathscr B}$-injective/projective object that is also $U$-injective/projective is also ${\mathscr A}$-projective/injective.

Answer 2

This is just a proof of the last thing you wrote.

Since $\Gamma$ is $\Lambda$-projective, $\Lambda^n \simeq \Gamma \oplus \Gamma'$ for some $n$ (possibly infinite). Thus considering $\Gamma$ as a sub-module of $\Lambda^n$, $i\otimes \Gamma$ is the restriction to $A'\otimes_\Lambda \Gamma$ of the map $i \otimes A^n: A'\otimes_\Lambda \Lambda^n \rightarrow A\otimes_\Lambda \Lambda^n$. This map is naturally identified with $i^{\oplus n}: (A')^n \rightarrow A^n$, which is injective if $i$ is. Hence its restriction to $A' \otimes_\Lambda \Gamma$ is also injective.