Question on the proof of the spectral mapping theorem for polynomials Theorem 13.9 in these lecture notes is the polynomial spectral mapping theorem:

Theorem 13.9 For a polynomial $p$ we have $\sigma(p(T)) = p(\sigma(T))$.


My questions:


*

*First direction: Why would $q(T) (p(T) - p(\lambda) I)$ be an inverse for $T - \lambda I$?
Update: @RobertIsrael said that instead $(p(T) - p(\lambda) I)^{-1} q(T)$ should be $(p(T) - p(\lambda) I)^{-1}$ be?
According to the lines in the proof above, we have
$$
(p(T) - p(\lambda) I)^{-1}
= ((T - \lambda I) q(T))^{-1}
\overset{?!}{=} q(T)^{-1} (T - \lambda I)^{-1},
$$
but as $\lambda \in \sigma(T)$ by assumption, $(T - \lambda I)^{-1}$ doesn't exist, right?

*For the second direction: Why does $x_i \in \sigma(T)$ imply that $p(x_i) - \mu = 0$? I thought that $x_i \in \sigma(T)$ only implies that $T - x_i I$ is not boundedly invertible.
Since $\mu \in \sigma(p(T))$ we have that $p(T) - \mu I = c \prod_{k = 1}^{n} (T - x_i I)$ is not boundedly invertible but how does that help?
In both statements there seem to be a notion of "a product of operators is not invertible iff all factors aren't invertible", but how can that be true? Consider i.e. shift operators on a sequence spaces such that $ST = I$, i.e. the left and right shift operators, then the above mentioned notion doesn't hold, right? 
 A: (1). It should be: "for otherwise $(p(T)-p(\lambda) I)^{-1} q(T)$ would give a bounded inverse for $T - \lambda I$."
(2). $p(x_i) - \mu = 0$ by substituting $x = x_i$ in the equation $p(x) - \mu =c \prod_{i} (x - x_i)$
EDIT: Note that all polynomial functions of $T$ and their inverses (when they exist) commute.  Thus from $q(T) r(T) = r(T) q(T)$, if $r(T)^{-1}$ exists we multiply by $r(T)^{-1}$ on both sides to get $r(T)^{-1} q(T) = q(T) r(T)^{-1}$.
The point in (1) is that $p(x) - p(\lambda) = (x -\lambda) q(x)$ means
$p(T) - p(\lambda) I = (T - \lambda I) q(T) = q(T) (T - \lambda I)$, and
so if $(p(T) - p(\lambda) I)^{-1}$ existed we could multiply the equation by that  to get
$I = (T - \lambda I) q(T) (p(T) - p(\lambda) I)^{-1} = q(T) (p(T) - p(\lambda) I)^{-1} (T - \lambda I)$, i.e. $q(T) (p(T) - p(\lambda)I)^{-1}$ is an inverse for $T - \lambda I$.
A: I will use the following theorems.

Let $g: A \to B$ and $f: B \to C$ be functions.
$(a)$ If $f \circ g$ is one-to-one, then $g$ is one-to-one.
$(b)$ If $f \circ g$ is onto, then $f$ is onto.

(Bounded Inverse Theorem): Let $X, Y$ be Banach spaces and assume $ T: X\to Y$ is a one-to-one, onto
  continuous linear operator.  Then $T^{−1}: Y\to X$ is continuous.

Also, $T$ is said to be bijective if $T$ is one-to-one and onto, and  $T$ is said to be invertible if $T$ is a bounded bijective linear map with a bounded inverse. So the Bounded Inverse Theorem states that every bounded bijective linear map from a Banach space to a Banach space is invertible.
From what I understand, here we are considering a Hilbert space $X$, which is a Banach space. So we can apply the Bounded Inverse Theorem. I would argue as follows.

It is easy to see that $p(T) - p(\lambda)I$ and $(T-\lambda I)$ are bounded linear operators(assuming $T$ is a bounded linear operator).We'll show that $p(T) - p(\lambda)I$ is not invertible.
On the contrary, assume $p(T) - p(\lambda)I$ is invertible. In particular, $p(T) - p(\lambda)I$ is bijective.  Then by the relation $$p(T) - p(\lambda)I=(T - \lambda I) q(T)= q(T)(T - \lambda I),$$ we see that $(T - \lambda I)$ is bijective(why?). Then by the Bounded Inverse Theorem, $(T - \lambda I)$ has a bounded inverse, a contradiction. 
So $p(T) - p(\lambda)I$ is not invertible. Hence $p(\lambda)\in\sigma(p(T))$.

As you can see, I didn't find out the inverse of $(T - \lambda I)$ explicitly. Existence of the bounded inverse itself is a contradiction and is sufficient for our purposes. 
For your second question, the equality $p(x_i) - \mu = 0$ follows from the factorisation of $p(t)-\mu$.
