# Equivalency of functions/graphs

by parsifal
Tags: equivalency, functions or graphs
 P: 15 I'm not sure where I'm supposed to put his, but I guess that at the core of my problem is the coordinate function, which I've seen in algebra related courses. I need some help with the concept of equivalency of functions or their graphs. I need to know if two functions or their graphs are equivalent at a point m ( f1(0)=f2(0)=m ). And I know the following about the situation: Graphs f1 and f2 on manifold M are equivalent at a point m (m is in an open group U), if for some chart fc(q1,...,qn): U -> Rn of manifold M $$\frac{d}{dt}q^i(f_1(t))=\frac{d}{dt}q^i(f_2(t))\ |_{t=0}$$ I also know that q is a coordinate function: $$q^i:=pr^i \circ f_c$$ where the projection $$pr^i:R^n \rightarrow R, (x^1,...,x^n) \mapsto x^i$$ I'm also told that if $$q^i:=pr^i \circ f_c$$ then f can be written as: $$f_c=(q^1,...,q^n)$$ How can fc be defined using the coordinate function q, if the coordinate function q was defined using the function fc? Also, I do not know what the projection function looks like, I only know about its mapping. Based on this info I should be able to find out if the two graphs f1 and f2 are equivalent at the point m. I can see how graphs can be considered equivalent, but I do not understand the meaning of the coordinate function in all this. And I suppose that f1 and f2 are charts of M, and therefore the coordinate function can be written as: $$q^i:=pr^i \circ f_1$$ Addition: After defining the equivalency using: $$\frac{d}{dt}q^i(f_1(t))=\frac{d}{dt}q^i(f_2(t))\ |_{t=0}$$ the material states that by using the chain rule it can be seen that this is an equivalence relation and doesn't depend on the chosen chart. If I use the chain rule I get $$\frac{dq^i}{df_1} \frac{df_1}{dt} = \frac{dq^i}{df_2} \frac{df_2}{dt}$$ and I do not see how the chart chosen does not matter. I'd appreciate the help. (I posted this on other forum earlier. I'm posting this here too, in the hope that someone here could help me. I hope you don't take me as a hard-core spammer.)

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