Rudin functional analysis, theorem 4.9 part (b)

Question

Let $M$ a closed subspace of a Banach space $X$.

b) Let $\pi:X \to X/M$ be the quotient map. Put $Y = X/M$, For each $y^* \in Y^*$, define $$ \tau y^* = y^* \pi $$ Then $\tau$ is an isometric isomorphism of $Y^*$.

Partial proof below:

If $x \ in X, y^* \in Y^*$, then $\pi x \in Y$; hence $x \to y^*\pi x$ is a continuous linear functional on $X$ which vanishes for $x \in M$. Thus $\tau y^* \in M^{\perp}$. The linearity of $\tau$ is obvious. Fix $x^* \in M^{\perp}$. Let $N$ be the null space of $x^*$. Since $M \subset N$ there's a linear functional $\Lambda$ on $Y$ such that $\Lambda \pi = x^*$.

Here's the question, which there's such a functional?

Answer 1

Take $y\in Y$, there exists $x\in X$ such that $\pi(x_y)=y$, write $\Lambda(y)=x^*(x_y)$, if $\pi(x'_y)=y$, $x_y-x'_y\in M$, this implies that $x^*(x_y-x'_y)=0$ and $\Lambda$ is well defined.