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So I have been reading some differential geometry and they are talking about density's and they claim that a pull back of a density is a density but I only have a partial proof of why this is true.

Let $M$ and $N$ be smooth manifold and let $F:M\rightarrow N$ be a smooth map. If $\mu$ is a density defined on $TN$ then the pullback of $\mu$ is defined to be $$(F^{*}\mu)_{p}(v_{1},\dots,v_{N})=\mu_{F(p)}(dF_{p}(v_{1}),\dots,dF_{p}(v_{n})).$$

So to show that $F^{*}\mu$ is a density all that must be done it to show that if $T$ is a linear transform on the tangent space $T_{p}M$ then $(F^{*}\mu)_{p}(Tv_{1},\dots,Tv_{N})=|det T|(F^{*}\mu)_{p}(v_{1},\dots,v_{N}).$

I can do this if I assume that $dF$ is invertible if this is the case then $dF\circ T\circ dF^{-1} $ is a linear transform of $TN$ so we have that

$\displaystyle (F^{*}\mu)_{p}(Tv_{1},\dots,Tv_{N})=\mu_{F(p)}(dF_{p}(Tv_{1}),\dots,dF_{p}(Tv_{n}))\\=\mu_{F(p)}(dF_{p}\circ T\circ dF_{F(p)}^{-1} \circ dF_{p}(v_{1}),\dots,dF_{p}\circ T\circ dF_{F(p)}^{-1} \circ dF_{p}(Tv_{n}))\\ = |det ~dF_{p}\circ T\circ dF_{F(p)}^{-1}| F^{*}\mu(v_{1},\dots,v_{n})\\ = |det ~T| F^{*}\mu(v_{1},\dots,v_{n}).$

Now this only work if $dF$ is invertible any idea on how to prove this with assuming that?

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1  
You meant "prove this without assuming that?" ? –  Sergio Parreiras Dec 31 '13 at 19:17
    
Just look for "change of variables formula" and "random vector" on Google. –  Sergio Parreiras Dec 31 '13 at 19:23
3  
Try considering separately the cases where $dF_p$ is invertible and where it's not. –  Jack Lee Jan 1 at 5:30

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