How does a morphism in a distinguished triangle induce an element of $\operatorname{Ext}^{2}$

Question

On p. 260 of Deligne and Illusie's paper (https://eudml.org/doc/143480) it says:

Let $e(K)\in\operatorname{Ext}^{2}(H^{1}(K),H^{0}(K))$ be the class defined by the degree $1$ morphism in the distinguished triangle $$H^{0}(K)\longrightarrow K\longrightarrow H^{1}K[-1]\longrightarrow^{+1}$$

What is meant by "the class defined by the degree 1 morphism"? How does a degree-1 morphism in a distinguished triangle define an element of $\operatorname{Ext}^{2}$? Sorry if that question is stupid, I am only a beginner when it comes to triangulated categories. Can someone please explain it to me in as much detail as possible?

Answer 1

A morphism of degree 1 from $A$ to $B$ is by definition the same as a morphism $A \to B[1]$. In your case, this is an element in $$ \mathrm{Hom}(H^1(K)[-1],H^0(K)[1]) \cong \mathrm{Hom}(H^1(K),H^0(K)[2]) \cong \mathrm{Ext}^2(H^1(K),H^0(K)). $$