# Diffeomorphism Preserves Tangency (Do Carmo 2.4.25)

Suppose $$C_1$$ and $$C_2$$ are regular curves on a regular surface $$S$$. Suppose $$p$$ is a point in $$S$$ where $$C_1$$ and $$C_2$$ are tangent, then if $$\varphi:S\rightarrow S$$ is a diffeomorphism, prove that $$\varphi(C_1)$$ and $$\varphi(C_2)$$ are regular curves which are tangent at $$\varphi(p)$$.

How would I start this? What would I need to show?

• As a start, can you write out in formulas what it means for $C_1$ and $C_2$ to be tangent at $p$? – quarague Feb 11 at 10:55
• @quarague Do you mean they have the same tangent plane at $p$? – JB071098 Feb 11 at 10:56

1. First, we have to show that if $$\varphi:S_{1}\rightarrow S_{2}$$ is a diffeomorphism, and $$C$$ is a regular curve on $$S_{1}$$, then $$\varphi(C)$$ is a regular curve on $$S_{2}$$. To do so, let $$\alpha:I\rightarrow S_{1}$$ be a parametrization of $$C_{1}$$. Then $$\varphi\circ\alpha:I\rightarrow S_{2}$$ is a parametrization of $$C_{2}$$, and we have $$\tag{1}\label{1} (\varphi\circ\alpha)'(t)=d_{\alpha(t)}\varphi(\alpha'(t)).$$ You should check that this expression is nonzero. Can you see why this is the case?

2. Next, we have two curves $$C_{1}$$ and $$C_{2}$$ that are tangent at $$p$$. This means that their velocity vectors at $$p$$ are parallel, i.e. there are parametrizations $$\alpha_{1}$$ and $$\alpha_{2}$$ of $$C_{1}$$ resp. $$C_{2}$$ such that $$\alpha_{1}(0)=\alpha_{2}(0)=p$$ and $$\tag{2}\label{2} \alpha_{1}'(0)=c\alpha_{2}'(0).$$ To show that $$\varphi(C_{1})$$ and $$\varphi(C_{2})$$ are tangent at $$\varphi(p)$$, we have to show that their velocity vectors $$(\varphi\circ\alpha_{1})'(0)$$ and $$(\varphi\circ\alpha_{2})'(0)$$ are parallel. Can you see that this follows from \eqref{1} and \eqref{2}?

• Yes that pretty much answers my question thank you. – JB071098 Feb 11 at 12:03
• Just to clarify, the fact that $\varphi\circ \alpha$ is a parametrization allows one to assert that the differential is nonzero correct? – JB071098 Feb 11 at 12:12
• @JB071098 The expression (1) above is nonzero because $\alpha'(t)$ is nonzero (since $\alpha$ is regular) and $d_{\alpha(t)}\varphi$ is an isomorphism (since $\varphi$ is a diffeomorphism). – studiosus Feb 11 at 13:20