User:Pmdboi/scratch pad

08:52		hey CW, can we talk here, getting confused with that "join/part" thing 08:52	 	ok 08:52		thanks alot sir 08:52	 	no worries 08:52		you see part a) 08:53	 	right 08:53	 	let $$g = \Psi_x(f) = (f^{-1}(x), \tilde{f})$$ 08:53	 	i'm not sure this is the right way, but i'm going to try it :) 08:53		ok thanks 08:53	 	then let $$h = \Psi_x^{-1}(g)$$ 08:53		yeah 08:54		g o h? 08:54		and h o g? 08:54		show they give identity? 08:54	 	then $$h(f^{-1}(x)) = x$$, right? 08:54		hmm why is that? 08:55		oh 08:55		yes 08:55	 	by the definition of $$G(y)$$ 08:55		right 08:55		ok 08:55	 	$$h(x) = \tilde{f}(f^{-1}(x)) = f(x)$$ 08:56		right 08:56	 	and $$h(y) = \tilde{f}(y) = f(y)$$ for all the other ones 08:56	 	i think i might have messed up the first one 08:56		oh 08:58		the first case? 08:58	 	right 08:58	*	pmdboi thinks 08:59	<JeffMat>	oh 09:00	 	oh, wait, no, everything is good 09:00	 	since $$x = f(f^{-1}(x))$$ 09:01	 	that shows that, for the three cases ($$y = f^{-1}(x)$$, $$y = x$$, and otherwise), $$h(y) = f(y)$$ 09:01	 	therefore, $$h = f$$ 09:02	 	and since $$h = \Psi_x^{-1} (\Psi_x(f))$$, $$\Psi_x^{-1} \circ \Psi_x$$ is the identity on $$\mathfrak{S}(E)$$ 09:02	 	or something like that 09:04	<JeffMat>	hmm