Welcome to our community

Be a part of something great, join today!

Problem of the Week #16 - September 17th, 2012

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Here's this week's problem.

-----

Problem: Let $M_1$, $M_2$ and $N$ be smooth manifolds. Show that $F=(f_1,f_2) : N\rightarrow M_1\times M_2$ is a smooth map if and only if $f_i : N\rightarrow M_i,\,i=1,2$ are smooth maps.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
No one answered this week's problem. Here's my solution below.

Suppose that $(f_1,f_2): N\rightarrow M_1\times M_2$ is smooth. There for coordinate charts $(\mathcal{U},\psi)$ in $N$ and $(\mathcal{V}\times\mathcal{W}, \varphi_1\times\varphi_2)$ in $M_1\times M_2$, it follows that the transition map $(\varphi_1\times\varphi_2)\circ(f_1,f_2)\circ \psi^{-1}$ is smooth. But it follows that $(\varphi_1\times\varphi_2)\circ(f_1,f_2)\circ \psi^{-1} = (\varphi_1\circ f_1\circ \psi^{-1},\varphi_2\circ f_2\circ \psi^{-1})$ is smooth. But since the cartesian product of smooth maps is smooth, it follows that each component is smooth, i.e. $\varphi_i\circ f_i\circ \psi^{-1}$ is smooth. But since the transition map $\varphi_i\circ f_i\circ \psi^{-1}$ is smooth, then it follows that $f_i$ is smooth. Reversing the argument shows that if $f_1$ and $f_2$ smooth, then $(f_1,f_2)$ is smooth.

Q.E.D.
 
Status
Not open for further replies.