# A question about surjective operators

Let $( \mathbb{E}\, , \leq )$ be a Riesz (vector lattice) space and let $\| \cdot \|$ be a map from $\mathbb{E}$ to $[ 0, \infty]$ that, when restricted to $$\mathbb{B} := \{ f \in \mathbb{E} \, \colon \, \| f \| < + \infty \},$$ $\| \cdot \|$ becomes a complete Riesz (lattice) norm on $\mathbb{B}$.

In particular, $( \mathbb{B}\, , \| \cdot \| )$ is a Banach lattice contained in $\mathbb{E}$. We denote by $\mathbb{E}_+$ and $\mathbb{B}_+$ be the positive cones of $( \mathbb{E}\, , \leq )$ and $( \mathbb{B}\, , \| \cdot \| )$, respectively.

Let $\Phi$ be an injective map from $\mathbb{E}_+$ to itself, i.e., $$\Phi \, \colon \mathbb{E}_+ \to \mathbb{E}_+ .$$

The preimage of the positive cone $\mathbb{B}_+ \subset \mathbb{E}_+$ under $\Phi$ is the subset of $\mathbb{E}_+$ for which $$\Phi^{-1} (\mathbb{B}_+) := \{ f \in \mathbb{E}_+ \, \colon \, \Phi (\, f \,) \in \mathbb{B}_+ \}.$$

I am quite confused that whether $\Phi \left( \Phi^{-1} (\mathbb{B}_+) \right) = \mathbb{B}_+$ holds? Could anyone help me out please?

Thank you very much in advance!

## 1 Answer

This is only true if $\Phi$ is surjective.

Let $E_+= \mathbb R^n_+$. Then $\Phi$ defined by $\Phi(x) = x + e$ is injective and maps into $E_+$ for $e\ne0$ and $e\ge0$. However, it is not surjective, as $\Phi(x)\ge e$ for all $x\in E_+$.

• Thanks so much @daw. As you mentioned, if $\Phi (\Phi^{-1} ( \mathbb{B}_+))=\mathbb{B}_+$ holds, then the mapping $\Phi \colon \mathbb{E}_+ \to \mathbb{E}_+$ must be needed to be surjective. May I confirm that is my understanding correct please? Many thanks again:) – Paradiesvogel Apr 28 '18 at 9:29
• yes, it is correct – daw Apr 28 '18 at 12:52