connectedness is preserved under a continuous map
Theorem Suppose $f:X\to Y$ 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 |
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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 |