Let $X, Y$ be $2$-dimensional smooth manifolds without boundary, and $P$ be a disconnected $1$-dimensional boundaryless embedded submanifold of $Y$. Let $f\colon X\to Y$ be a smooth map that is transverse to $P$. So, $f^{-1}(P)$ is a $1$-dimensional boundaryless embedded submanifold of $X$.
Let $C$ be a component of $P$ and $C'$ be a component of $f^{-1}(C)$. Is $f\big |C'\to C$ a submersion?
My attempt: $f$ is transverse to $P$ implies $f$ is transverse to $C$ from definition.
Let $x\in C'$. So, $T_x(C')=T_x\big(f^{-1}(C)\big)=df_x^{-1}\big(T_{f(x)}(C)\big)$. The first equality is due to connectedness and the last equality due to transversality.
Therefore, $df_x\big(T_x(C')\big)=df_x\big(df_x^{-1}\big(T_{f(x)}(C)\big)\big)\subseteq T_{f(x)}(C)$.
Since $T_{f(x)}(C)$ is of $1$-dimensional vector space, the vector subspace $df_x\big(T_x(C')\big)$ can be either $0$ or $1$-dimensional.
Now, I have to show $df_x\big(T_x(C')\big)=T_{f(x)}(C)$.
No. Counterexample: $X = Y = \mathbb R^2$, $P = x$-axis and $$f : \mathbb R^2 \to \mathbb R^2, \ \ \ f(x, y) = (x^3, y).$$
Then $f$ is transverse to $P$, $f^{-1} (P) = x$-axis, but
$$f|_{f^{-1}\ (P)} (x, 0) = (x^3, 0)$$ is not a submersion.