# Injection of Ext groups

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