# Prove that $(\mathbf g\circ\mathbf f)_*=\mathbf g_*\circ\mathbf f_*$ and $(\mathbf g\circ\mathbf f)^*=\mathbf g^*\circ\mathbf f^*$

Let $$\mathbf f:\mathbf R^n\rightarrow\mathbf R^m$$ and $$\mathbf g:\mathbf R^m\rightarrow\mathbf R^k$$. I figured out the pull-back part by finding $$(\mathbf g\circ\mathbf f)^*(du_1)=d(g_1(f_1(x_1,...,x_n),...f_m(x_1,...,x_n))$$ I then let $$h_1=g_1(f_1,...,f_m)$$ and got $$(\mathbf g\circ\mathbf f)^*(du_1)=\sum_{i=1}^{n}\frac{\partial h_1}{\partial x_i}dx_i=\sum_{i=1}^{n}(\sum_{j=1}^{m}\frac{\partial g_1}{\partial y_j}*\frac{\partial f_i}{\partial x_i}dx_i)=\sum_{j=1}^{m}\frac{\partial g_1}{\partial y_j}*\sum_{i=1}^{n}\frac{\partial f_i}{\partial x_i}dx_i$$ I also found that the other side equals this for the pull-back half of the question so I think that was right. I am just not sure how to proceed for the push-forwards. Any help on that or comments at my attempt for this half would be much appreciated.

Sorry if that's not written as cleanly as it should be I'm still familarizing myself with MathJax.