Continuous image of a connected space

The proof is short. If \(f(X)\) could be written as the union of two disjoint nonempty open sets \(U\) and \(V\) of the subspace, then \(f^{-1}(U)\) and \(f^{-1}(V)\) would be disjoint nonempty open sets of \(X\) with union \(X\), contradicting the connectedness of \(X\).

On the real line the theorem becomes the intermediate value theorem. The connected subsets of \(\mathbb R\) are the intervals, so the image of an interval under a continuous function is another interval: if \(f(a)=-2\) and \(f(b)=3\), the image contains all of \([-2,3]\), in particular \(0\). For the same reason, \(x\mapsto 1/(1+x^2)\) maps \(\mathbb R\) onto \((0,1]\).

The theorem also works as a negative test: if a continuous surjection from a connected space onto \(Y\) exists, then \(Y\) is connected. Hence \(\mathbb R\setminus\{0\}\) is not a continuous image of \(\mathbb R\), and a continuous function from a connected space into a discrete space is constant.

Nothing else is guaranteed about the image: it may be neither open, nor closed, nor compact. The converse fails too: a constant map has a connected image even if its domain is not connected. As a consequence, connectedness is invariant under homeomorphisms, which makes it useful for telling spaces apart, and the idea of joining points by continuous curves leads to path-connectedness.