# Continuous injective map

Consider a topological space $X$ and a subset $Z \subset X$. Assume we are given a continuous injective map (in $X$ topology) $f:Z \to Z$ such that $g: Z \setminus C \to Z$(where $C \subset Z$) and $g\circ f = {\rm id}_Z$ is also continuous. Will $f$ and $g$ remain continuous in the induced topology, i.e. the topology induced on $Z$ by the topology of $X$?

Thank you

-

Since the domains of $f$ and $g$ are in $Z$, all the open preimages under them are in $Z$. Thus they remain open in the induced topology, since they don't change by intersection with $Z$. Thus $f$ and $g$ remain continuous.
@Ali: If you already know the answer and just want to make sure you're not making a mistake, it would be preferable to phrase the question accordingly ("It seems to me that $f$ and $g$ should remain continuous because ...; or am I making a mistake?"). The more information you provide on what you already know and have already done, the more specific the responses can be. Also if you've already done the work and come up with the answer, why make someone else do the same work and write it out? – joriki Oct 20 '11 at 12:14
But maybe care must be taken to the fact that $g^{-1} (U \cap Z) =g^{-1}(U) \cap g^{-1}(Z)=V \cap Z\setminus C$ which is not necessarily open ? (I took $U$ and $V:= g^{-1}(U)$ open sets in the $X$-topology). – user17090 Oct 20 '11 at 12:30
@Ali: $g^{-1}(U\cap Z)$ is not necessarily $g^{-1}(U)\cap g^{-1}(Z)$. There may be elements in $U$ and $Z$ with common preimages, and they may not lie in $U\cap Z$. – joriki Oct 21 '11 at 9:46