0
$\begingroup$

Let $\mathcal A \subset \mathcal B$ be an exact inclusion of a full abelian subcategory $\mathcal A$ into an abelian category $\mathcal B$. Assume, that both $\mathcal A$ and $\mathcal B$ have enough injectives (or enough projectives). Since the Ext functor classifies extensions, we have an inclusion $\mathrm{Ext}^i_{\mathcal A}(X,Y) \subset \mathrm{Ext}^i_{\mathcal B}(X,Y)$. The map can also be constructed via resolutions. Is there a way to show, that the map $\mathrm{Ext}^i_{\mathcal A}(X,Y) \to \mathrm{Ext}^i_{\mathcal B}(X,Y)$ is injective without any reference to extensions?

A standard example is the base restriction functor $K\text{-}\mathrm{Vect} \to K[x]\text{-}\mathrm{Mod}$ along $K[x] \to K, x \mapsto 0$, where $K$ is a field. Then $\mathrm{Ext}^1_{K}(K,K)=0$ and $\mathrm{Ext}^1_{K[x]}(K,K)=K$.

I don't want to assume, that the inclusion preserves injectives or projectives. In this case any resolution in $\mathcal A$ computes the derived functor in $\mathcal B$ and the map is bijective.

The claim is false, when the functor $\mathcal A \to \mathcal B$ is faithful, but not full.

This question is related: Exact functor and relationship between Ext functors

$\endgroup$
4
$\begingroup$

I don't think the claim is correct: Consider e.g. a field $k$ and the full inclusion of $k[x]/(x^2)\text{-mod}$ into $k[x]\text{-mod}$. This cannot induce an injection on extension groups because $k[x]/(x^2)$ is of infinite global dimension, while $k[x]$ has global dimension $1$.

The problem with the attempt to reason in terms of Yoneda extensions is that $\text{Ext}^k(X,Y)$ are the connected components of the directed graph of length-$k$ extensions, and when considered in ${\mathscr B}$ there may be paths between extensions living in ${\mathscr A}$ which pass through intermediate extensions which are outside of ${\mathscr A}$. This cannot happen for $k=1$ because of the Five lemma. So, broadly speaking the problem is simply that $A\subset B$ doesn't imply $\pi_0(A)\subset\pi_0(B)$.

$\endgroup$
1
  • $\begingroup$ @JulianQuast : Do you have any questions? $\endgroup$ – Hanno Dec 14 '20 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.