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.

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.

Bibliography

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.