# The differentials of compositions of $C^{\infty}$ maps obey a chain rule.

$M,N,L$ are $n_{1},n_{2},n_{3}$-differentiable manifolds respectively, $m\in M$, $f:M\rightarrow N$ and $g:N\rightarrow L$ are $C^{\infty}$, and let $J_{f}(m)$ denotes the Jacobian of $f$ at $m$. I could prove $(gof)_{*}|_{m}=g_{*}|_{f(m)}of_{*}|_{m}$ but I couldn't complete that $J_{gof}(m)=J_{g}(f(m))\times J_{f}(m)$.

My works is below:

$J_{gof}(m)=\big[\frac{‎\partial‎(z_{i}(gof))}{‎\partial‎ x_{j}}|_{m}\big]_{n_{3}\times n_{1}}$

$J_{g}(f(m))\times J_{f}(m)=\big[\frac{‎\partial‎(z_{i}og)}{‎\partial‎ y_{j}}|_{f(m)}\big]_{n_{3}\times n_{2}}\times\big[\frac{‎\partial‎(y_{i}of)}{‎\partial‎ x_{j}}|_{m}\big]_{n_{2}\times n_{1}}$ $=\big[\sum_{k=1}^{n_{2}}(\frac{‎\partial‎(z_{i}og)}{‎\partial‎ y_{k}}|_{f(m)}\cdot\frac{‎\partial‎(y_{k}of)}{‎\partial‎ x_{j}}|_{m})\big]_{n_{3}\times n_{1}}$

$\frac{‎\partial‎(z_{i}og)}{‎\partial‎ y_{i}}|_{f(m)}\cdot\frac{‎\partial‎(y_{i}of)}{‎\partial‎ x_{i}}|_{m}=g_{*}|_{f(m)}(\frac{\partial z_{i}}{\partial y_{k}})\cdot f_{*}|_{m}(\frac{\partial y_{k}}{\partial x_{j}})$

I think perhaps I should show that $\sum_{k=1}^{n_{2}}\big(g_{*}|_{f(m)}(\frac{\partial z_{i}}{\partial y_{k}})\cdot f_{*}|_{m}(\frac{\partial y_{k}}{\partial x_{j}})\big)=(g_{*}|_{f(m)}of_{*}|_{m})(\frac{\partial z_{i}}{\partial x_{j}})$ that it is $(gof)_{*}|_{m}(\frac{\partial z_{i}}{\partial x_{j}})$ from the first part.

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Composition is denoted by \circ, not o. – Zhen Lin Dec 15 '12 at 15:25

If you believe $(g \circ f)_*|_m = g_*|_{f(m)} \circ f_*|_m$, and want to know that $J_{g \circ f}(m) = J_g(f(m)) \times J_f(m)$, then the question is pretty much why matrix multiplication is the same as composition of linear transformations, which you can check directly!