I am teaching myself representation theory on $GL(V)$ and $S_n$ using my friend's lecture notes, and have reached a proof of the Schur-Weyl Duality theorem; on reading through I'm struggling to make my way through the first part of the proof though, and was hoping you might be able to explain a few of the things I'm confused about. I'm not sure how much of this proof is standard (a fair amount, I suspect) so it's possible some questions aren't going to be answerable and will simply be the nuances of the way the lecturer has presented the material. However, if you could help me as much as you can with my understanding I would greatly appreciate it. (I will write out the first part of this section of the notes but I won't present the full proof in its entirety.)
A forewarning: the lecture notes have quite a few errors in; the last step of the proof of Schur-Weyl duality for example is totally wrong, though was easily fixed via earlier results in the notes. Also, please forgive any stupid questions I ask, almost all of my representation theory knowledge is self-taught, so I occasionally overlook the blindingly obvious. My questions are I think simple, although I have quite a bit that needs clarifying.
Let $V$ be an $m$-dimensional vector space over $\mathbb{C}$. $V^{\otimes n}$ is a $\mathbb{C}S_n$ module, via $\sigma(v_1 \otimes \ldots \otimes v_n) = v_{\sigma^{-1}(1)} \otimes \ldots \otimes v_{\sigma^{-1}(n)}$, and $g(v_1 \otimes \ldots \otimes v_n) = gv_1 \otimes \ldots \otimes gv_n$ (I presume this is for $g \in GL(V)$ though not explicitly stated.)
We can define $\phi: \mathbb{C}S_n \to \operatorname{End}_\mathbb{C}(V^{\otimes n}),\,\sigma \mapsto (v \mapsto \sigma v)$. $V$ can also be regarded as representations of $GL(V)$ in a natural way, so $V^{\otimes n}$ becomes a $\mathbb{C}GL(V)$ module, so $\exists \, \psi: \mathbb{C}GL(V) \to \operatorname{End}_{\mathbb{C}GL(V)}(V^{\otimes n}), \eta \mapsto \eta^{\otimes n}$.
Theorem (Schur-Weyl Duality): The images of $\mathbb{C}S_n$ and $\mathbb{C}GL(V)$ in $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ are each others centralizers; that is, $\psi(GL(V)) = \operatorname{End}_{\mathbb{C}S_n}(V^{\otimes n})$ and $\phi(\mathbb{C}S_n) = \operatorname{End}_{\mathbb{C}GL(V)}(V^{\otimes n})$.
In order to prove this, we define the $\mathbb{C}$-algebra $S_\mathbb{C}(m,n)$, the "Schur algebra", to be the subalgebra of $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ consisting of endomorphisms commuting with the image of $\mathbb{C}S_n$: that is , $S_\mathbb{C}(m,n) = \operatorname{End}_{\mathbb{C}S_n}(V^{\otimes n})$. Now $\mathbb{C}S_n$ is semisimple and $V^{\otimes n}$ is a finite dimensional $\mathbb{C}S_n$ module, therefore $S(m,n)$ is a semisimple $\mathbb{C}$-algebra. Put $W = \operatorname{End}_\mathbb{C}(V)$: this is a $\mathbb{C}GL(V)$ module by conjugation.
Question 1: Is there a reason why we would want a $\mathbb{C}GL(V)$ module by conjugation? I can't see any point during any of the proof at which conjugation is used, and I wouldn't have thought this would be the natural action to choose.
Lemma: There exists an isomorphism $\alpha: W^{\otimes n} \to \operatorname{End}_\mathbb{C}(V^{\otimes n})$, sending $f_1 \otimes \ldots \otimes f_n \mapsto \left(v_1 \otimes \ldots \otimes v_n \mapsto f_1(v_1) \otimes \ldots \otimes f_n(v_n)\right)$, an isomorphism of $\mathbb{C}GL(V)$-modules and of $\mathbb{C}S_n$-modules.
Proof: $W^{\otimes n} = W \otimes \ldots \otimes W \cong (V \otimes V^*) \otimes \ldots \otimes (V \otimes V^*)$ $ \cong (V \otimes V \ldots \otimes V) \otimes (V^* \otimes \ldots \otimes V^*) \cong V^{\otimes n} \otimes (V^*)^{\otimes n} \cong \operatorname{End}_\mathbb{C}(V^{\otimes n})$.
$W^{\otimes n}$ has natural structure of a $\mathbb{C}S_n$-module and $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ inherits its structure from $V^{\otimes n}$. It is easy to see $\alpha$ is a $\mathbb{C}S_n$-homom; this completes the lemma.
Question 2: I follow the symbolic part of the proof, and I guess $W^{\otimes n} = (\operatorname{End}_\mathbb{C}(V))^{\otimes n}$ has the "natural" structure of a $\mathbb{C}S_n$ module via $\sigma(f_1 \otimes \ldots \otimes f_n) = f_{\sigma^{-1}(1)} \otimes \ldots \otimes f_{\sigma^{-1}(n)}$? I don't know what is meant by $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ inheriting its structure from $V^{\otimes n}$: does this mean $\sigma$ acts by mapping $\left(v_1 \otimes \ldots \otimes v_n \mapsto f_1(v_1) \otimes \ldots \otimes f_n(v_n)\right)$ to $\left(v_1 \otimes \ldots \otimes v_n \mapsto f_{\sigma^{-1}(1)}\left(v_{\sigma^{-1}(1)}\right) \otimes \ldots \otimes f_{\sigma^{-1}(n)}\left(v_{\sigma^{-1}(n)}\right)\right)$ in the same way as was stated in the first line?
If this is correct however, then I can't see how $\alpha$ commutes with $\sigma$: the latter has indices consistent between $f_i$ and $v_i$ whereas the former has $f_{\sigma^{-1}(i)}(v_i)$. But then how is $\alpha$ a $\mathbb{C}S_n$-homomorphism? What have I misunderstood? Back to the notes:
Lemma: $S_\mathbb{C}(m,n) = \alpha(T^nW_{sym})$, where $T^nW_{sym}$ denotes the n-th symmetric power.
Proof: Observe $S_\mathbb{C}(m,n) = \{x \in \operatorname{End}_\mathbb{C}(V^{\otimes n}): \sigma x = x \, \forall \, \sigma \in S_n\}$, and $T^nW_{sym} = \{y \in W^{\otimes n}: \sigma y = y \, \forall \, \sigma \in S_n\}$, completing the proof.
Question 3: I'm probably being stupid, but why does $S_\mathbb{C}(m,n) = \{x \in \operatorname{End}_\mathbb{C}(V^{\otimes n}): \sigma x = x \, \forall \, \sigma \in S_n\}$? It's meant to be endomorphisms commuting with the image of $\mathbb{C}S_n$, but wouldn't that give something like $\sigma x = x \sigma$, rather than $\sigma x = x$?
That's all for now; apologies for the lengthy/multiple questions, but I didn't see any benefit to posting the same lecture notes 3 times in separate questions. Thank you in advance for your help.