Question 3.1.1 in AT

Question

Here is the question I am trying to understand its solution:

Show that $\operatorname{Ext}(H, G)$ is a contravariant functor of $H$ for fixed $G$, and a covariant functor of $G$ for fixed $H.$

Here is a solution I found online:

Here are my questions:

1- why $H^1(F,G) = \operatorname{Ext}(H,G),$ is this by the universal coefficient theorem?

2- Why the author is calling $H^1(F,G)$ the homology group of the dual at the end of the paragraph that begins with the word "suppose"? should not it be called cohomology? He is also doing this in the last paragraph.

3- Where is actually the proof that $(\beta \alpha)^* = \alpha* \beta*$ in the solution? he did not say why we reversed the order.Also, why in the last paragraph $(\beta \alpha)_*$ does not reverse the order? Is there a commutative diagram for proving so?

Could anyone help me answer these questions please?

Answer 1

1. It's the definition. Hatcher says on p. 195, "Thus the only interesting group $H^n(F;G)$ is $H^1(F;G)$. As we have seen, this group depends only on $H$ and $G$, and the standard notation for it is $\mathrm{Ext}(H, G)$."

2. Sure. But homology, cohomology, it doesn't really matter what you call it. Start with a (co)chain complex and compute kernels mod images.

3. If we have $H \xrightarrow{\alpha} H' \xrightarrow{\beta} H''$, then we have the composite $(\beta \alpha): H \to H''$. Now apply $\mathrm{Hom}(-, G)$ to get $(\beta \alpha)^* = \alpha^* \beta^*$ on Hom. Now extend both of these maps on the chain cocomplexes: extend $(\beta \alpha)^*$ and extend each map separately. By the uniqueness part of Lemma 3.1, they induce chain homotopic maps on the cochain complexes, and hence they induce the same map on cohomology: on Ext we also have $(\beta \alpha)^* = \alpha^* \beta^*$.