Sections of associated bundles Let $\pi:P\rightarrow M$ be a Principal bundle and $\pi_V:P\times_G F\rightarrow M$ be its associated bundle via the representation $\rho:G\rightarrow GL(V)$.
Fact: 
$\Gamma(P\times_G V)\simeq\{f:P\rightarrow V: f(pg)=\rho(g^{-1})f(p)\}=:A$
I am trying to get a better understanding of this statement. 
Suppose $f\in A$. We want to show that this $f$ corresponds to a unique $\sigma\in \Gamma(P\times_G V)$. So unless I'm mistaken, given $f$, we want to construct a map $\sigma:M\rightarrow P\times_G V$ such that $\pi_V\circ \sigma=\text{id}$.
Based on lecture notes found here, I think the first step in this process is to define the map
$F:P\rightarrow P\times_G V$ 
by $F=[\text{id}\times f]$.
Then $F(pg)=[pg,\rho(g^{-1})f(p)]=[p,f(p)]=F(p)$
The lecture notes then say this map 'descends' to a section $\sigma:M\rightarrow P\times_G V$. 
I am wondering how to construct this 'descent'? Is it done by defining $\sigma=F\circ \pi^{-1}$? Does this make sense? It seems to be 'well defined' since we just take any $x\in M$ to its fiber which is then mapped uniquely to an equivalence class. Just seems a bit dodgy as $\pi^{-1}$ isn't a function per say.
Conversely, if $\sigma\in \Gamma(P\times_G V)$, then $\sigma$ 'lifts' to a map $F:P\rightarrow P\times_G V$ and since $\pi_F\circ \sigma=\text{id}$, it follows that $F=\text{id}\times f$. I can't seem to get my head around this explanation of the converse argument and I am hoping someone is able to help me.
 A: After many hours, I believe I have answered my own question:
Claim:
There is a 1-1 correspondence
\begin{align*}
\{f:P\rightarrow V: f(pg)=\rho(g^{-1})f(p) \}\simeq \Gamma(P\times_G V).
\end{align*}
Proof:
Let $f:P\rightarrow V$ be as described. Define local sections $\tilde{s}_{\alpha}:U_{\alpha}\rightarrow P\times_G V$ by
\begin{align*}
\tilde{s}_{\alpha}(m)&=[(s_{\alpha}(m),f(s_{\alpha}(m))]
\end{align*}
Define $\sigma=\tilde{s}_{\alpha}$ on $U_{\alpha}$. Then $\sigma$ is a global section since the definition agrees on overlaps. Indeed, for $m\in U_{\alpha}\cap U_{\beta}$,
\begin{align*}
\tilde{s}_{\beta}(m)&=[(s_{\beta}(m),f(s_{\beta}(m))]\\
&=[(s_{\alpha}(m)g_{\alpha\beta}(m),f(s_{\alpha}(m)g_{\alpha\beta}(m))]\\
&=[(s_{\alpha}(m)g_{\alpha\beta}(m),\rho(g_{\beta\alpha}(m))f(s_{\alpha}(m))]\\
&=[((s_{\alpha}(m),f(s_{\alpha}(m))]\\
&=\tilde{s}_{\alpha}(m)
\end{align*}
So each $f$ corresponds to an element in $\Gamma(P\times_G V)$.
Conversely, suppose $\sigma\in \Gamma(P\times_G V)$. The section $\sigma$ allows us to construct a function $f_{\sigma}:P\rightarrow V$ in the following way: Let $p\in P$ be abritrary and write $m=\pi(p)$. We define the function $f_{\sigma}$ as
\begin{align*}
f_{\sigma}(p)&=v,\quad\text{such that } \sigma(m)=[(p,v)].
\end{align*}
This is a well defined construction since if we choose another representative of $P\times_G V$ and write $\sigma(m)=[(q,w)]$ for some $q\in P$ and $w\in V$, we have that $q=pg$ for some $g\in G$ ($p$ and $q$ both lie in $\pi^{-1}(m)$), and then just set $v=\rho(g)w$. 
This constructed $f_{\sigma}$ satisfies the desired property. Indeed, suppose $f(pg)=\tilde{v}$. That is, $\tilde{v}$ is the vector such that $\sigma(\pi(pg))=\sigma(m)=[(pg,\tilde{v})]$. Then we have
\begin{align*}
\sigma(m)=[(pg,\tilde{v})]&=[(p,\rho(g)\tilde{v})].
\end{align*}
In other words, $f(p)=\rho(g)\tilde{v}=\rho(g)f(pg)$.
Would anyone be able to verify if that is correct?
