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$A,B$ arbitrary sets, $f_i : A \to B$, $i \in \{1,...,n\}$ functions. Now let $f_j$ be non invertable for a $j \in \{1,...,n\}$. Does this already imply that $(\sum_{i=1}^n f_i)(x)$ is non invertable?

Pazu
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1 Answers1

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No, simple counterexample. Let $f_j \colon \mathbb{R} \mapsto \mathbb{R}$ be given by $f_j(x)=x^j$, for $j=1,2,3$. Then (as you can check) $(f_1+f_2+f_3)(x)$ is everywhere increasing, so invertible, while $f_2$ is clearly not invertible.

kjetil b halvorsen
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