Corollary 2

Let X and Y be topological spaces and let $f: X\rightarrow$ Y be a continuous mapping. In order that $f$ be relatively connected, it is necessary and sufficient that the following three properties be satisfied:

a) The image $f(X)$ is dense in Y;

b) The space Y is locally connected; $-_1$

c) For every open and connected set V of Y, the set $f(V)$ is connected.

Suppose that the mapping $f$ is relatively connected. The density of $f(X)$ in Y follows from the definition of a relatively

connected mapping. Let V be an open subset of Y; the mapping $f_V:\overset{-1}{f}(V)\rightarrow$ V is relatively connected. By Prop. 4, the connected components of V are open and closed in V. It follows that Y is locally connected (TG, I, p. 85, prop. 11). To prove assertion c), it is enough to prove that X is connected if Y is. By the lemma, there exists

a connected component U of X such that $Y =\overline{f(U)}$ and $\overset{-1}{f}(f(U)) = U$, whence U = X and X is connected.

Conversely, suppose that conditions a), b), c) are satisfied, and let us show that $f$ is relatively connected. Let $y$ be a point of Y; since Y is locally connected, $y$ has a fundamental system of connected open neighbourhoods. If W is such a neighbourhood, conditions

c) and a) imply that $\overset{-1}{f}(W)$ is connected and nonempty. Hence the mapping $f$ is relatively connected.