Question about the Poincaré first return function Let $M$ be a compact and connected two-dimensional differentiable manifold of class $C^2$. Let $\varphi: \mathbb R \times M \to M$ be a flow of class $C^2$ in $M$. Let $\sigma:[-1,1]\to M$ be an embedding of class $C^2$ such that $\Sigma:=\sigma((-1,1))$ is transversal to every orbit of $\varphi$. Let $V\subset [-1,1]$ be the set of points whose image has its first return in $\Sigma$.. We define the real function $f:V\to (-1,1)$ as follows $v\in (-1,1)$ is carried by $\sigma$ to $\sigma(v )\in\Sigma$ then the first return to $\Sigma$ from $\sigma(v)\in\Sigma$ is $\varphi(t_{\sigma(v)},\sigma(v))$ now it we return to $(-1,1)$ by $\sigma^{-1}$. So that
$$
f(v):=\sigma^{-1}(\varphi(t_{\sigma(v)},\sigma(v)))
$$

I'm trying to prove that $f'(v)\neq 0$ for all $v$, but I can't see it. My ideas were to prove that $f$ is strictly monotone but I failed in the attempt. My second attempt was to get an explicit relationship between $f'(v)$ with $\sigma$ and $\varphi$ but I've been having a hard time getting anywhere. Maybe the answer is obvious but I need something concrete to justify it because I don't see it so clearly. Any suggestion will be of incredible help, please detail your answer a bit so I can understand it.
 A: Fried talks about first return maps in this paper: https://www.sciencedirect.com/science/article/pii/0040938382900179
Here's the basic idea. Let $M$ be a compact and connected two-dimensional differentiable manifold of class $C^2$. Let $\varphi: \mathbb R \times M \to M$ be a flow of class $C^2$ in $M$. Let $\sigma:[-1,1]\to M$ be an embedding of class $C^2$ such that $\Sigma:=\sigma([-1,1])$ is transversal to every orbit of $\varphi$ (notice a slight change to ensure $\Sigma$ is closed). We wish to define a first return time map $g : \Sigma \rightarrow (0,\infty)$. For $x \in \Sigma$, define
$$ T(x) := \{t \in (0,\infty) \ | \ \varphi^t(x) \in \Sigma\}.$$
Define $g$ by
$$g(x) := \inf\{t > 0 \ | \ \varphi^t(x) \in \Sigma\}.$$
Since $\Sigma$ is transversal to every orbit of $\varphi$, we see that the set is non-empty, so an infimum exists and this defines a function. We need to check that this function is nice.
Claim: This infimum is achieved, so $g$ is a well-defined function $g : \Sigma \rightarrow (0,\infty)$.
Proof:  Let $(t_n) \subseteq T(x)$ be such that $t_n \rightarrow g(x)$. This means that for each $n$ we have $\varphi^{t_n}(x) \in \Sigma$. Since $\Sigma$ is compact, we can find a subsequence $(n_k)$ so that $\varphi^{t_{n_k}}(x) \rightarrow y \in \Sigma$. Since $\varphi$ is $C^2$, we have $\varphi^{g(x)}(x) = y \in \Sigma$, so the infimum is achieved. $\blacksquare$
Claim: The function $g(x)$ is $C^2$.
Proof: By taking local coordinates around $x$ and $\varphi^{g(x)}(x)$, we can use the implicit function theorem to guarantee a $C^2$ map $\hat{g}(x)$ in a neighborhood of $x$. Furthermore, on this neighborhood we have that $\hat{g} = g$.(*) $\blacksquare$
In our set up, $V$ will just be $[-1,1]$, so there's no point in distinguishing them. As a consequence, consider the map $f : [-1,1] \rightarrow [-1,1]$ defined by $f(t) := \sigma^{-1}(\varphi^{g(\sigma(t))}(\sigma(t))$ (this is the first return map). This map is $C^2$ and admits a $C^2$ inverse defined by flowing backwards. As a consequence, we cannot have $f'(t) = 0$.
(*) Note: this is a little dubious and should be checked. Generally people skip over how I defined $g$ and just say that $\hat{g}$ is the first return map so they don't have to worry about this. I know of an argument to show that $g$ is continuous without using the implicit function theorem, but I'd be interested if someone could show that $g$ is $C^1$ without the implicit function theorem.
