Lemma 62.13.1. The construction above is bilinear, i.e., we have $(\alpha _1 + \alpha _2) \circ \beta \alpha _1 \circ \beta + \alpha _1 \circ \beta $ and $\alpha \circ (\beta _1 + \beta _2) = \alpha \circ \beta _1 + \alpha \circ \beta _2$.
Proof. Omitted. Hint: on fibres the construction is bilinear by Lemma 62.11.1. $\square$
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