connectedness is preserved under a continuous map

TheoremSuppose $f:X\to Y$ is a continuous mapbetween 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 $Y=(0,1)\cup (2,3)$ showsthat 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 thereare 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.