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Let $\mathcal{C}$ be an abelian category (feel free to put more adjectives here) and $\mathcal{D}$ a full abelian subcategory closed under kernels and cokernels.

Then by definition for $A,B\in \mathcal{D}$, $\mathrm{Hom}_{\mathcal{D}}(A,B)\simeq \mathrm{Hom}_{\mathcal{C}}(A,B)$.

Moreover, since the subcategory is full, an extension $0\to B\to X\to A\to 0$ with $X\in \mathcal{D}$ splits in $\mathcal{D}$ if and only if it splits in $\mathcal{C}$, so we have that $$\mathrm{Ext}^1_{\mathcal{D}}(A,B)\subseteq \mathrm{Ext}^1_{\mathcal{C}}(A,B).$$

Can we somehow argue this way and conclude that $\mathrm{Ext}^i_{\mathcal{D}}(A,B)\subseteq \mathrm{Ext}^i_{\mathcal{C}}(A,B)$ for all $i$?

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  • $\begingroup$ No, it does not hold. For example, take for C (resp. D) the category of modules over the ring Z (resp. Z/4), where Z is the ring of integers. Ext vanishes from degree 2 in C, but is zero in no positive degree in D. $\endgroup$ Commented May 18, 2021 at 5:27

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