@redtree Maybe you could quote which part of Woit's book you don't understand.
Woit speaks about linear representations ##(V,\pi)## on complex vector spaces ##V##, i.e. complex-linear mappings
\begin{align*}
\pi\, : \,G&\longrightarrow \operatorname{GL}(V) \\
A&\longmapsto (v\longmapsto A(v)=A\cdot v)
\end{align*}
with ##\pi(A\cdot B)=\pi(A)\cdot\pi(B)##. ##A,B## are group elements and ##\pi(A),\pi(B)## are matrices. If we choose ##G## to be a matrix group itself, e.g. ##U(1)## or ##U(n)## then we already have matrices and we can assume that ##\pi=\operatorname{id}_G,## i.e. ##\pi ## doesn't do anything.
If he says unitary representation, then he requires that the vector space ##V## has an inner product such that
\begin{align*}
\bigl\langle x\, , \,ay+z \bigr\rangle&=a\bigl\langle x\, , \,y \bigr\rangle +\bigl\langle x\, , \,z \bigr\rangle \\
\bigl\langle ax+y\, , \,z \bigr\rangle &=\overline{a}\bigl\langle x\, ,\, z \bigr\rangle + \bigl\langle y\, , \,z \bigr\rangle \tag{1} \\
\bigl\langle x\, , \,y \bigr\rangle &=\overline{\bigl\langle y\, , \,x \bigr\rangle } \\[12pt]
\bigl\langle \pi(A)(v)\, , \,\pi(A)(w) \bigr\rangle &=\bigl\langle v\, , \,w \bigr\rangle
\end{align*}
If ##A\in U(n)## is a unitary matrix, then
$$
\bigl\langle A(v)\, , \,A(w) \bigr\rangle =\bigl\langle A^\dagger A(v)\, , \,w \bigr\rangle =\bigl\langle I(v)\, , \,w \bigr\rangle =\bigl\langle v\, , \,w \bigr\rangle \tag{2},
$$
means the natural representation (of unitary matrices on a complex vector space with a (Hermitian) inner product) is a unitary representation.
We don't have any fancy group homomorphisms, the representations are all linear and finite-dimensional, and the vector spaces are complex with a complex inner product, means a Hermitian inner product because of semi-linearity ##(1).## Unitary matrices lead to unitary representations by simply letting them act on complex vectors.
Woit writes ##\pi ## since the theorems hold for any linear, complex, finite-dimensional representation, some even for infinite-dimensional representations. Still, you can understand the matter by thinking of ##\pi## being the identity:
\begin{align*}
\pi = \operatorname{id}_{U(n)}\, : \,U(n)&\longrightarrow \operatorname{GL}(\mathbb{C}^n) \\
A&\longmapsto (v\longmapsto A(v)=A\cdot v)\\[12pt]
\begin{pmatrix}a_{11}&\ldots & a_{1n}\\ \vdots &&\vdots \\ a_{n1}&\ldots & a_{nn}\end{pmatrix}&\longmapsto \left(\begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}\longmapsto \begin{pmatrix}a_{11}&\ldots & a_{1n}\\ \vdots &&\vdots \\ a_{n1}&\ldots & a_{nn}\end{pmatrix}\cdot \begin{pmatrix}x_1\\ \vdots \\ x_n\end{pmatrix}\right)
\end{align*}
If you want to be very picky, then the only difference between elements of ##U(n)## and their representation as a matrix in ##\operatorname{GL}(n,\mathbb{C})=\operatorname{GL}(\mathbb{C}^n)## is, that the matrices ##A## on the LHS, the elements of ##A\in U(n)## are just group elements which happened to be defined as matrices, and their homomorphic images ##\pi(A)=\operatorname{id}_{U(n)}(A)=A \in \operatorname{GL}(\mathbb{C}^n)## on the RHS are explicitly seen as linear transformations that map vectors from ##\mathbb{C}^n## onto vectors from ##\mathbb{C}^n.## But this is a very, very academic way to distinguish them. And such a perspective becomes obsolete anyway the moment we directly apply ##A## on ##v## and write ##A(v)## without referring to a representation like I did (out of laziness) in ##(2).## Formally, it should have been
\begin{align*}
\bigl\langle \pi(A)(v)\, , \,\pi(A)(w) \bigr\rangle &=\bigl\langle \operatorname{id}(A)(v)\, , \,\operatorname{id}(A)(w) \bigr\rangle = \bigl\langle A(v),A(w) \bigr\rangle \\
&=\bigl\langle A^\dagger (A(v))\, , \,w \bigr\rangle =\bigl\langle (A^\dagger A)(v)\, , \,w \bigr\rangle
=\bigl\langle I(v)\, , \,w \bigr\rangle \\
&=\bigl\langle v\, , \,w \bigr\rangle
\end{align*}
