Proof check-Adjoint of an operator

Question

I'm working on this problem:

Let $M$ be a closed subspace of a Banach space $X.$ Deline $\epsilon \colon M \to X$ as $\epsilon(x) = x$ for $x \in M.$ Prove the adjoint $\epsilon^* \colon X^* \to M^*$ is defined by $\epsilon^*\mu = \mu|_M$ for $\mu \in X^*.$

My proof:

$\epsilon^{*} \mu=\mu\epsilon$ where $\mu\epsilon: M \to \mathbf{F}$. Since $\mu(\epsilon(x))=\mu(x)$, thus $\mu\epsilon=\mu|_{M}$.

Therefore $\epsilon^{*} \mu=\mu \epsilon=\left.\mu\right|_{M}$ for every $\mu \in X^*$.

It looks relatively short, did I miss anything?

Answer 1

Everything looks good. The proof works.