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Let $\mathcal{C}$ be a full subcategory of $\text{Mod}_A$, the category of modules over a commutative unital ring $A$ (I am mostly interested in the graded setting, but I doubt there is much difference here between graded and ungraded cases).

Let $I$ be an injective $A$-module contained in $\mathcal{C}$. Must $I$ be an injective object in $\mathcal{C}$? Let $M$ and $N$ be modules in $\mathcal{C}$, $\iota:M\to N$ be a monomorphism in $\mathcal{C}$, and $\alpha:M\to I$ be any morphism. If $\iota$ is also a monomorphism in $\text{Mod}_A$, then this gives a morphism $\beta:N\to I$ in $\mathcal{C}$ with $\beta\iota=\alpha$, so that $I$ is an injective object in $\mathcal{C}$.

But how do we know that $\iota$ is a monomorphism in $\text{Mod}_A$? Let $X$ be any $A$-module and $\varphi,\psi:X\to M$ with $\iota\varphi=\iota\psi$. Need to show that $\varphi=\psi$. But we only know from the fact that $\iota$ is mono in $\mathcal{C}$ that this works for $X\in\mathcal{C}$. Am I missing something?

EDIT: In the case I am most interested in, we have the following property: if $M, N\in\mathcal{C}$ and $f:M\to N$, then $\text{ker}(f)\in\mathcal{C}$. So in the argument above, $\text{ker}(\iota)\in\mathcal{C}$. Let $\varphi$ be the inclusion $\text{ker}(\iota)\to M$ and $\psi$ be zero, so that $\iota\varphi=\iota\psi$. Since $\iota$ is mono in $\mathcal{C}$, we know that $\varphi=\psi$, completing the proof.

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2 Answers 2

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The following is a counterexample: take $A = \mathbb{Z}$ and $\mathcal{C}$ be the full subcategory containing $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$. Then I claim the natural quotient map $q: \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z}$ is a monomorphism. Granted the claim, it is clear then $\mathbb{Q}$ cannot be injective in $\mathcal{C}$ despite being an injective abelian group, as the identity morphism $\mathbb{Q} \rightarrow \mathbb{Q}$ cannot be extended to $\mathbb{Q}/\mathbb{Z}$.

Now, to prove the claim, we note that the only morphism from $\mathbb{Q}/\mathbb{Z}$ to $\mathbb{Q}$ is the zero map, so there’s nothing to verify there. Thus, it suffices to show that, if $f, g: \mathbb{Q} \rightarrow \mathbb{Q}$ are morphisms s.t. $q \circ f = q \circ g$, then $f = g$. We observe that $q \circ f = q \circ g$ means $q \circ (f - g) = 0$, whence $f - g$ is a morphism from $\mathbb{Q}$ to $\mathbb{Z}$. But the only such morphism is the zero map, so $f = g$.

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  • $\begingroup$ This is in line with Sergey’s answer, since $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ are the first two terms of an injective resolution of $\mathbb{Z}$. $\endgroup$
    – David Gao
    Commented Dec 14, 2023 at 0:48
  • $\begingroup$ In fact, since for every nonzero element $z$ of $\mathbb{Z}$ there is some nonzero integer $n$ s.t. $z$ cannot be divided by $n$, any morphism from any injective abelian group to $\mathbb{Z}$ is the zero map, so the above argument shows that $\mathbb{Q}$ is not injective even in the full subcategory of all injective abelian groups. Furthermore, the same argument also tells you that in the full subcategory of all injective abelian groups, the only injective object is the trivial group. (Since if it’s nontrivial you can embed either $\mathbb{Z}$ or some $\mathbb{Z}/n$ into it.) $\endgroup$
    – David Gao
    Commented Dec 14, 2023 at 1:34
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The answer should be 'No'. I will first show this is the case if we replace injective objects with projective objects.

Let $A=\mathbb{Z}$ and $\mathcal{C}$ the full subcategory of finitely-generated free abelian groups. Then the morphism $\mathbb{Z}\xrightarrow{2}\mathbb{Z}$ is an epimorphism in this category. If $\mathbb{Z}$ were projective in $\mathcal{C}$, it would admit a section, which it clearly does not.

I suppose you could build a counter-example for the injective case by proceeding analogously: take $\mathcal{C}$ the category of injective modules over $A$, consider some non-injective module $M$ and the beginning of its injective resolution $I^0\to I^1$. This is a morphism that has a nontrivial kernel in the whole category of modules, but I expect it to be a monomorphism in the full subcategory of injective modules, though I will have to think about it a bit more. If this is indeed the case, we find that $I^0$ can't be injective in $\mathcal{C}$, since then we would be able to truncate the resolution to a single term resolution $I^0$ and we would have $M\cong I^0$.

A useful sufficient condition for the projectives/injectives to behave nicely under restriction to a subcategory is the exactness of the inclusion functor.

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    $\begingroup$ I believe you need stronger assumptions on $M$ for this to work, namely that every morphism from an injective module to $M$ is the zero map. Since given a nonzero morphism $I \rightarrow M$, composing it with the inclusion $M \rightarrow I^0$ results in a map that is nonzero but composes with $I^0 \rightarrow I^1$ to the zero map. $\endgroup$
    – David Gao
    Commented Dec 14, 2023 at 1:27

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