Let $X$ be a Banach space, $u: X \to X$ be compact and $\lambda$ a non zero complex number. Let $W=X/(u-\lambda)(X)$. Let $\pi : X \to W$ be the quotient map. If $X^\ast$ denotes the dual I assume that $\pi^\ast$ denotes the quotient map
$$ \pi^\ast : X^\ast \to X^\ast / (u^\ast-\lambda)(X^\ast)$$
Could someone please help me understand why $\mathrm{im}(\pi^\ast)\subseteq \mathrm{ker}(u^\ast - \lambda)$? It should be "obvious" but it's not obvious to me. What is obvious to me is that $\mathrm{im}(u^\ast -\lambda) \subseteq \ker{\pi^\ast}$. Maybe I am misunderstanding what the map $\pi^\ast$? This is all mentioned on page 22 in Murphy's book.
Edit The following is an image of the relevant page:
Your assumption is not correct. The notation $\pi^*$ is standard and refers to the dual map $\pi^*:W^*\to X^*$ given by $\pi^*(f)=f\circ \pi$.
Now $\sigma$ is a bounded functional on $X$. Because it is zero on the subspace $(u-\lambda)X=\ker\pi$, one can define $\tau(\pi(x)):=\sigma(x) $. This is well-defined: if $\pi(x)=\pi(y) $, then $x-y\in\ker\pi$, so $\sigma(x-y)=0$ and $\sigma(x)=\sigma(y) $. Finally, $\tau$ is bounded because, for any $v\in\ker\pi$, $$ |\tau(\pi(x))|=|\sigma(x+v)|\leq\|\sigma\|\,\|x+v\|, $$ so $$ |\tau(\pi(x))|=\leq\|\sigma\|\,\inf\{\|x+v\|:\ v\in\ker\pi\}=\|\sigma\|\,\|\pi(x)\|. $$