I am confused about the definition of orientation on manifolds.
Let $X=\{(x,y,0)\in \mathbb{R}^3 \mid x^2+y^2=1\}$ and $Y=\{(x,y,1)\in \mathbb{R}^3 \mid x^2+y^2=1\}$ be two one dimensional circles in $\mathbb{R}^3$ with different hight ($z$-values).
Suppose $X$ and $Y$ has the counterclockwise orientation.
Let $f_1$ be a map given by $(x,y,z)\mapsto (x,y,z+1)$ and let $f_2$ to be a map given by $(x,y,z)\mapsto (x,y,1-z)$.
In $\mathbb{R}^3$, $f_1$ is just a shift and $f_2$ is a reflection. So $f_1$ is orientation preserving homeomorphism of $\mathbb{R}^3$ to itself and $f_2$ is orientation reversing.
However, when we restrict $f_1$ and $f_2$ to $X$ then both become the map sending $(x,y,0)$ to $(x, y,1)$, hence $f_1=f_2$ on $X$.
Is this orientation preserving map from $X$ to $Y$?
I am really confused about the definition of orientation and the question itself might be nonsense. Please guide me.
The confusion arises because the two $1$-dimensional manifolds are embedded in a two dimensional manifold (namely the cylinder with equation $x^2+y^2=1)$. Denote the orientations on the cicles as the 1-forms $\mathrm{d}\theta_X$ and $\mathrm{d}\theta_Y$. Now the cylinder has an orientation defined by $\mathrm{d}\theta\wedge\mathrm{d}z$, from which the two 1-forms are restrictions to $X$ and $Y$. Now for the function $f_2$ we have $f_2^*(\mathrm{d}\theta\wedge\mathrm{d}z)=f_2^*(\mathrm{d}\theta)\wedge f_2^*(\mathrm{d}z)=\mathrm{d}\theta\wedge(-\mathrm{d}z)=-\mathrm{d}\theta\wedge\mathrm{d}z$. So we see that $f_2$ reverses the orientation of the cylinder but preseves that of $\mathrm{d}\theta$ and reverses $\mathrm{d}z$. If you can show that exactly the opposite happens for the function $f_3(x,y,z)=(x,-y,-z)$ then you have fully understood this answer.