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?
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2$\begingroup$ 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 $\endgroup$ – Maxime Ramzi May 17 '19 at 20:22
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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.
answered Jun 21 '19 at 5:37
