# $\operatorname{Ext}^0$ for free resolutions

I am studying homological algebra for an exam in algebraic topology, and I was wondering:

Let $H,G$ be two abelian groups. What is $\operatorname{Ext}^0(H;G)$?

Now here's what I have done:

We can always take a free resolution of $H$ of the following form: $$0\to F_1\stackrel{f_1}{\longrightarrow}F_0\stackrel{f_0}{\longrightarrow}H\to0$$ (where $F_1$ represents the relations of $H$, and $F_0$ its generators after a choice of basis $\{b_i\}_i$). Dualizing we have the cochain complex: $$0\leftarrow F_1^*\stackrel{f_1^*}{\longleftarrow}F_0^*\stackrel{f_0^*}{\longleftarrow}H^*\leftarrow0$$ where $H^*=\hom_\mathbb{Z}(H,G)$ etc. An element of $H^*$ is a homomorphism $\phi:H\to G$, which is completely determined by the image of the basis elements $b_i$. The homomorphism $f_0^*\phi:F_0\to G$ is then given by $f_0^*\phi(b_i)=\phi(b_i)$. Thus $f_0^*$ is injective and: $$\operatorname{Ext}^0(H;G)=\frac{\ker f_0^*}{\operatorname{im}(0\to H^*)}=0$$ Is this correct?

• How do you dualise and maintain the property of being exact? – Zhen Lin Jun 20 '14 at 22:41
• @ZhenLin I don't. The original sequence is exact, the second is not. That's why we get non trivial $\operatorname{Ext}^n$. However in the computation above I did something wrong (namely, I computed the homology at the wrong level, I think). – Daniel Robert-Nicoud Jun 23 '14 at 1:14

If $F : \mathcal{C} \to \mathcal{D}$ is any left exact functor between abelian categories, then $R^0 F = F$. Because if $0 \to A \to I^0 \to I^1 \to \dotsc$ is an injective resolution, then $0 \to F(A) \to F(I_0) \to F(I_1)$ is exact, i.e. $(R^0 F)(A) = \ker(F(I_0) \to F(I_1))=F(A)$.

In particular, for $\mathcal{D}=\mathsf{Ab}$ and $F=\mathrm{Hom}(X,-)$ we get $\mathrm{Ext}^0(X,Y)=\mathrm{Hom}(X,Y)$.

If you want to use projective resolutions for computing $\mathrm{Ext}$: Just dualize the above statement, so that for every right exact functor $G : \mathcal{C} \to \mathcal{D}$ we have $L^0 G = G$. Applying this to the right exact Hom functor $\mathrm{Hom}(-,Y) : \mathcal{C} \to \mathsf{Ab}^{\mathrm{op}}$, we get again $\mathrm{Ext}^0(X,Y)=\mathrm{Hom}(X,Y)$.

Apply $\DeclareMathOperator{Hom}{Hom}\Hom(H,-)$ to an injective resolution $$0\to G\xrightarrow{u} I^0\xrightarrow{i} I^1\to\dotsb$$ to obtain a complex $$0\to\Hom(H,I^0)\xrightarrow{f}\Hom(H,I^1)\to\dotsb$$ where $f(\phi)=i\circ\phi$. By definition, $\DeclareMathOperator{Ext}{Ext}\Ext^0(H,G)=\ker f$.

We claim that if $A\xrightarrow{\ell}I^0$ with $i\circ\ell=0$, then there exists a unique $A\xrightarrow{k_\ell} G$ such that $u\circ k_\ell=\ell$. Indeed, if $A\xrightarrow{\ell}I^0$ satisfies $i\circ\ell=0$, then putting $k_\ell(a)=u^{-1}(\{\ell(a)\})$ works (check this!).

Now, define $$\Phi:\ker f\to\Hom(H,G)$$ by $\Phi(\phi)=k_\phi$. Then $\Phi$ is an isomorphism (check this!). Hence $\Ext^0(H,G)\simeq\Hom(H,G)$.

• Ok, I see... So, other than the fact that you are using an injective resolution instead of a free resolution, and that your $G$ is my $H$ and vice versa, you are saying that I computed the homology at the wrong degree? – Daniel Robert-Nicoud Jun 19 '14 at 21:22
• @Karl I couldn't care less. The downvote is enough--no need to bother me with an inane comment. – Brian Fitzpatrick Jun 20 '14 at 21:47
• @Karl See my above comment. – Brian Fitzpatrick Jun 20 '14 at 21:51
• @Karl Also caps are pretty unchill dude. – Brian Fitzpatrick Jun 20 '14 at 21:51
• @Karl I don't understand why you don't just post another answer instead of spewing this vitriol. We come to MSE for positive vibes not negative ones. – Brian Fitzpatrick Jun 20 '14 at 22:00