# connectedness is preserved under a continuous map

The inclusion map  for spaces $X=(0,1)$ and $Y=(0,1)\cup(2,3)$ shows that we need to assume that the map is surjective. Othewise, we can only prove that $f(X)$ is connected. See this page (http://planetmath.org/IfFcolonXtoYIsContinuousThenFcolonXtoFXIsContinuous).

Proof. For a contradiction   , suppose there are disjoint open sets $A,B$ in $Y$ such that $Y=A\cup B$. By continuity and properties of the inverse image, $f^{-1}(A)$ and $f^{-1}(B)$ are open disjoint sets in $X$. Since $f$ is surjective, $Y=f(X)=A\cup B$, whence

 $X=f^{-1}f(X)=f^{-1}(A)\cup f^{-1}(B)$

contradicting the assumption  that $X$ is connected.

## References

• 1
• 2 G.L. Naber, Topological methods in Euclidean spaces, Cambridge University Press, 1980.
Title connectedness is preserved under a continuous map ConnectednessIsPreservedUnderAContinuousMap 2013-03-22 13:55:59 2013-03-22 13:55:59 drini (3) drini (3) 7 drini (3) Theorem msc 54D05 CompactnessIsPreservedUnderAContinuousMap ProofOfGeneralizedIntermediateValueTheorem