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


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 $\pi_0(A)\subset\pi_0(B)$.

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

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