connectedness is preserved under a continuous map

Theorem Suppose $f\colon X\to Y$ is a continuous map between topological spaces $X$ and $Y$ . If $X$ is a connected space, and $f$ is surjective, then $Y$ is a connected space.

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 G.J. Jameson, Topology and Normed Spaces, Chapman and Hall, 1974.
2 G.L. Naber, Topological methods in Euclidean spaces, Cambridge University Press, 1980.

Title	connectedness is preserved under a continuous map
Canonical name	ConnectednessIsPreservedUnderAContinuousMap
Date of creation	2013-03-22 13:55:59
Last modified on	2013-03-22 13:55:59
Owner	drini (3)
Last modified by	drini (3)
Numerical id	7
Author	drini (3)
Entry type	Theorem
Classification	msc 54D05
Related topic	CompactnessIsPreservedUnderAContinuousMap
Related topic	ProofOfGeneralizedIntermediateValueTheorem