# Exact functor and relationship between Ext functors

Let $$F:\mathcal{A}\to\mathcal{B}$$ be an exact functor betwen two abelian categories. Let $$r\geq 1$$ be an integer. I wonder what is the relationship between $$Ext^r_{\mathcal{A}}(X,Y)$$ and $$Ext^r_{\mathcal{B}}(F(X),F(Y))$$. As far as I understand, an exact functor preserves long exact sequences, and if two long exact sequences in $$\mathcal{A}$$ are equivalent, then the fact that the functor $$F$$ preserves composition and identities implies that the images of the two sequences will also be equivalent in $$\mathcal{B}$$. Hence elements of $$Ext^r_{\mathcal{A}}(X,Y)$$ give rise to elements of $$Ext^r_{\mathcal{B}}(F(X),F(Y))$$ but of course there could be more extensions in $$\mathcal{B}$$ between $$F(X)$$ and $$F(Y)$$. This should imply that $$Ext^r_{\mathcal{A}}(X,Y) \subseteq Ext^r_{\mathcal{B}}(F(X),F(Y))$$. Is my argument correct?

• Well two sequences could become equivalent in $\mathcal{B}$ even if they weren't in $\mathcal{A}$ so you don't have an inclusion per se – Maxime Ramzi May 17 '19 at 20:22

What your argument gives is a homomorphism $$\operatorname{Ext}^r_\mathcal{A}(X,Y)\to\operatorname{Ext}^r_{\mathcal{B}}(F(X),F(Y))$$ (to see that it is a homomorphism, just note that applying $$F$$ to everything preserves the Baer sum operation since $$F$$ is exact). However, this homomorphism may not be injective, so there is not necessarily any reasonable way you can consider $$\operatorname{Ext}^r_\mathcal{A}(X,Y)$$ as a subset of $$\operatorname{Ext}^r_{\mathcal{B}}(F(X),F(Y))$$. For instance, $$F$$ might be the zero functor in which case $$\operatorname{Ext}^r_{\mathcal{B}}(F(X),F(Y))$$ is always trivial, regardless of whether $$\operatorname{Ext}^r_\mathcal{A}(X,Y)$$ was trivial.