# Show that is a Morse function.

Let $$f:M\to\mathbb{R}$$ a Morse function, and $$\pi:\tilde{M}\to M$$ a covering map, show that $$f\circ \pi:\tilde{M}\to \mathbb{R}$$ is also a Morse function.

Let $$\tilde{f}=f\circ \pi$$, then $$d\tilde{f}_{\tilde{p}}=d(f\circ \pi)_{\tilde{p}}=df_{\pi(\tilde{p})}\circ d\pi_{\tilde{p}}=df_p\circ d\pi_{\tilde{p} } .$$ Hence $$\begin{array}{lll} \tilde{p}\in Cr(\tilde{f})&\iff & d\tilde{f}_{\tilde{p}}=0 \\ &\iff &df_p\circ d\pi_{\tilde{p} }=0\\ \end{array}$$

We know that $$\pi$$ is a local diffeomorphism, so $$d\pi_{\tilde{p}}=0$$ then, i will get $$df_p=0$$, i.e. $$p\in Cr(f)$$. Therefore. $$\boxed{Cr(\tilde{f})=\{\tilde{p}\in\tilde{M}|p\in Cr(f)\}.}$$

It just remains to show that if $$\tilde{p}\in Cr(\tilde{f})$$ then it is non-degenerated. How can I do this?

Thank you.

EDITED

If $$\tilde{p}\in Cr(\tilde{f})$$, then $$p\in Cr(f)$$ and we have

$$\begin{array}{lll} d^2\tilde{f}_{\tilde{p}}(v_1,v_2)&= &d(df_p\circ d\pi_{\tilde{p}})\\ \\ &= &d^2f_{p}(d\pi_{\tilde{p}}(v_1),d\pi_{\tilde{p}}(v_2))+df_{p}(d^2\pi_{\tilde{p}}(v_1,v_2))\\ \\ &= &d^2f_{p}(d\pi_{\tilde{p}}(v_1),d\pi_{\tilde{p}}(v_2))\quad\mbox{ Using that df_p=0.} \end{array}$$

Knowing that $$f$$ is a Morse function, then $$p$$ is non-degenerated, then , $$\tilde{p}$$ is also non-degenerate. Therefore, $$\tilde{f}=f\circ \pi$$ is a Morse function.

I know that the second detivative is only defined at critical points, but I also used the chain rule, is that ok? Thank you.

• Do you mean $d\pi$ is invertible? – Keshav Nov 30 '19 at 20:31
• Since $\pi$ is a local diffeomorphism, you only need to show that the Hessian is well-defined at critical points. You can do this in coordinates with the chain rule, or appeal to bilinear form definition for a slicker proof. – Elliot G Nov 30 '19 at 20:37
• @Elliot I will try that, thank you. – Framate Nov 30 '19 at 20:44
• @ElliotG, I added what I tried, would you mind to take a look? Thank you. – Framate Dec 8 '19 at 5:49