# composition of mappings

• Feb 16th 2011, 11:02 PM
Zennie
composition of mappings
Let $F: R^n \rightarrow R^m$ and $G: R^m \rightarrow R^p$ be mappings. Prove that their composition $GF: R^n \rightarrow R^p$ is a differentiable mapping. (Take $m=p=2$ for simplicity). Then show that $(GF)_* = G_*F_*$.

I do not know how to prove this and would appreciate some help. Thanks!
• Feb 17th 2011, 04:59 AM
HallsofIvy
Quote:

Originally Posted by Zennie
Let $F: R^n \rightarrow R^m$ and $G: R^m \rightarrow R^p$ be mappings. Prove that their composition $GF: R^n \rightarrow R^p$ is a differentiable mapping. (Take $m=p=2$ for simplicity). Then show that $(GF)_* = G_*F_*$.

I do not know how to prove this and would appreciate some help. Thanks!

The first thing you should know is that you can't prove it. What you have stated is untrue. The composition of two arbitrary mappings is NOT in general a differentiable mapping.

Did you mean to say that F and G are themselves differentiable mappings? What is your definiton of "differentiable"?