Transition maps for principal bundle are smooth

Let \(\pi:P\rightarrow M\) be a principal \(G\) bundle. We choose an open covering \(\{U_\alpha\}\) of \(M\) and trivializations \(\psi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times G\) defined as \(\psi_\alpha(u)= (\pi(u),\varphi_\alpha(u))\) such that \(\varphi_\alpha(ua)=\varphi_\alpha(u)a\) for all \(u\in \pi^{-1}(U_\alpha)\) and \(a\in G\). Let \(x\in U_\alpha\cap U_\beta\). Given \(v\in \pi^{-1}(x)\subseteq \pi^{-1}(U_\alpha)\cap \pi^{-1}(U_\beta)\), we have \(\varphi_\alpha(v)\in G\) and \(\varphi_\beta(v)\in G\). For \(v'\in \pi^{-1}(x)\) there exists \(g\in G\) such that \(v'=vg\). Then, we have

\(\varphi_\alpha(v')\varphi_\beta(v')^{-1}= \varphi_\alpha(vg)\varphi_\beta(ua)^{-1} =\varphi_\alpha(u)aa^{-1}\varphi_\beta(u)^{-1} =\varphi_\alpha(u)\varphi_\beta(u)^{-1}\)

Thus, for any \(v,v'\in \pi^{-1}(x)\), we have

\(\varphi_\alpha(v)\varphi_\beta(v)^{-1}=\varphi_\alpha(v')\varphi_\beta(v')^{-1}.\)

This gives a well defined map \(g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G\) given by  \(g_{\alpha\beta}(x)=\varphi_\alpha(u)\varphi_\beta(u)^{-1}\) where \(u\in P\) is such that \(\pi(u)=x\). Now, I have to check that this map \(g_{\alpha\beta}\) is smooth. As \(\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times G\) is smooth, so is its projection to \(G\) i.e., the map \(\pi^{-1}(U_\alpha)\rightarrow G\) given by \(u\mapsto \varphi_\alpha(u)\) is smooth. Similarly the map \(\pi^{-1}(U_\beta)\rightarrow G\) given by \(u\mapsto \varphi_\beta(u)\) is smooth. As inverse map on \(G\) is smooth, so is the composition \(\pi^{-1}(U_\beta)\rightarrow G\rightarrow G\) with \(u\mapsto \varphi_{\beta}(u)^{-1}\). So, the map \(\pi^{-1}(U_\alpha\cap U_\beta)\rightarrow G\times G\) given by \(u\mapsto (\varphi_{\alpha}(u),\varphi_{\beta}^{-1}(u))\) is smooth. As multiplication map \(G\times G\rightarrow G\) is smooth, so is the compostion \(\pi^{-1}(U_\alpha\cap U_\beta)\rightarrow G\times G\rightarrow G\) given by \(u\mapsto \varphi_{\alpha}(u)\varphi_\beta^{-1}(u)\). Thus, the map \(\pi^{-1}(U_\alpha\cap U_\beta)\rightarrow G\) given by \(u\mapsto \varphi_{\alpha}(u)\varphi_{\alpha}\beta^{-1}(u)\) is smooth. Proving \(g_{\alpha\beta}:U_\alpha\cap U_\beta\rightarrow G\) is smooth boils down to proving \(U_\alpha\cap U_\beta\rightarrow \pi^{-1}(U_\alpha\cap U_\beta)\) is smooth. As \(\pi^{-1}(U_\alpha)\rightarrow U_\alpha\times G\) is diffeomorphism, its inverse  \(U_\alpha\times G\rightarrow \pi^{-1}(U_\alpha)\) is smooth and so is its composition with inclusion \(U_\alpha\rightarrow U_\alpha\times G\) given by \(x\mapsto (x,1)\). Thus, we have smooth map \(U_\alpha\rightarrow \pi^{-1}(U)\) given by \(x\mapsto (x,1)\) and so is the map \(U_\alpha\cap U_\beta\rightarrow \pi^{-1}(U_\alpha\cap U_\beta)\). Thus, the map

\(U_\alpha\cap U_\beta\rightarrow \pi^{-1}(U_\alpha\cap U_\beta)\rightarrow G\)

which is precisely \(g_{\alpha\beta}\) is smooth.