This is a question I encountered in a topology class： Let f : (R,$T_{st}$) → (R,$T_{st}$) be a continuous function (Here, $T_{st}$ is the standard topology defined as := {(a, b) | a < b}).
1)Is the function G$|^{G(\rm I\!R)}$:($\rm I\!R,T_{st}$) → (G($\rm I\!R)$,${T_{st}}_{G (\rm I\!R)}$), obtained by restricting the codomain of G to its image, continuous?
2) Is the function G$|^{G(\rm I\!R)}$:($\rm I\!R,T_{st}$) → (G($\rm I\!R)$,${T_{st}}_{G (\rm I\!R)}$ a homeomorphism?
My thoughts: for (1), let G(x) = [x] where [x] represents the greatest integer like n $\leq$ x$<$n+1. And I take (3/2, 5/2) in standard topology, n $\cap$ (3/2, 5/2) = {2} which is a singleton, and is open in the subspace topology of the standard topology on N. Am I on the right track? If not, could you guys give some idea how to do this?
Thoughts on (2): I think the map is not injective (1 to 1), but I don't know how to prove it or find a counter example.
Thank you in advance for the help!
In general, if you have a topological space $(X,\tau_{X})$ and $A\subseteq X$, then the subspace topology, $(A,\tau_{A})$, is the finest topology such that the inclusion map, $\iota \colon (A,\tau_{A})\longrightarrow (X,\tau_{X})$, is continuous.
Moreover, if $(X,\tau_{X})$, $(Y,\tau_{Y})$ are topological spaces and $B\subseteq Y$, $f\colon (X,\tau_{X})\longrightarrow (B,\tau_{B})$ is continuous if and only if $\iota \circ f \colon (X,\tau_{X})\longrightarrow (Y,\tau_{Y})$ is continuous.
With respect to the second point, let $f\colon (\mathbb{R},\tau_{U})\longrightarrow (\mathbb{R},\tau_{U})$ be such that $f(x)=x_{0}$ for some fixed $x_{0}\in \mathbb{R}$. Then, $f$ is continuous, but $\mathbb{R}$ obviously is not homeomorphic to $\{x_{0}\}$.