Proposition 6

The mapping $(f,g) \mapsto \operatorname{Ext}_A(f,g) :$

is $k$-bilinear.

Let $f\in\operatorname{Hom}_A(M',M)$, $g_1,g_2\in\operatorname{Hom}_A(N,N')$, $\lambda_1,\lambda_2\in k$. The morphisms $\operatorname{Homgr}_A(L(f),\lambda_1g_1+\lambda_2g_2)$ and $\lambda_1\operatorname{Homgr}_A(L(f),g_1)+\lambda_2\operatorname{Homgr}_A(L(f),g_2)$ from $\operatorname{Homgr}_A(L(M),N)$ to $\operatorname{Homgr}_A(L(M),N')$ coincide, hence, by Proposition 5 and Remark 3,

(10) $\operatorname{Ext}_A(f,\lambda_1g_1+\lambda_2g_2)=\lambda_1\operatorname{Ext}_A(f,g_1)+\lambda_2\operatorname{Ext}_A(f,g_2)$.

Argue similarly for the mapping $f \mapsto \operatorname{Ext}_A(f,g)$.