Appendix 446 A Proof of Proposition 1 in Section 2 447 Proof

ReLU (T (v u) + b) = ReLU( Tv + b), where u = 0, that is, ReLU (T + b) is not injective. By injectivity of T, we finally get a = b . Remark 2. An example that satisfies (3.1) is the neural operator whose This construction is given by the combination of "Pairs of projections" discussed in Kato [2013, Section I.4.6] with the idea presented in [Puthawala et al., 2022b, Lemma 29]. R. We write operator null G by Thus, in both cases, H is injective. Remark 4. W e make the following observations using Theorem 1: Leaky ReLU is one of example that satisfies (ii) in Theorem 1. Puthawala et al. [2022a, Theorem 15] assumes that We first revisit layerwise injectivity and bijectivity in the case of the finite rank approximation.