Proposition 4

Let X and Y be topological spaces and let $f: X\rightarrow Y$ be a relatively connected continuous mapping.

a) For every connected component U of X$,\overline{f(U)}$ is a

connected, open and closed subset of Y, and one has $\overset{-1}{f}(f(U)) = U$.

b) For every connected component V of Y, there exists a connected component U of X such that $V =f(U)$. The mapping of U into V induced by $f$ by passing to the subsets is relatively connected.

c) The connected components of X ( resp. of Y) are open and closed.

a) Let V be the set of $y\in Y$ which possess a neighbourhood W such that

$\overset{-1}{f}(W)$ is connected and meets U; this is an open set of Y. It contains $\overline{f(U)}$ since $f$ is relatively connected. Conversely, let $y\in V$; let

W be a neighbourhood of $y$ such that $\overset{-1}{f}(W)$ is connected and meets U. For

every neighbourhood $W'$ of $y$ such that $\overset{-1}{f}(W')$ is connected, $\overset{-1}{f}(W\cap W')$ is

not empty, hence $\overset{-1}{f}(W\cup W')$ is connected (TG, I, p. 81, prop. 2). By

hypothesis, $\overset{-1}{f}(W\cup W')$ meets U; hence one has $\overset{-1}{f}(W\cup W')\subset U$ and, a fortiori, $W'\cap f(U)=\not\emptyset$. This proves that $y\in \overline{f(U)}$; therefore, $V =\overline{f(U)}$. In particular, the set $\overline{f(U)}$ is open and closed in Y; prop. 1 (TG, I, p. 81) further implies that it is connected.

The preceding arguments show moreover that $\overset{-1}{f}(f(U))\subset U$,

whence the equality $\overset{-1}{f}(f(U)) = U$, the other inclusion being obvious. In particular, U is open and closed in X.

b) Let V be the connected component of a point $y$ of Y, let W be a

neighbourhood of $y$ such that $\overset{-1}{f}(W)$ is connected and nonempty and let U be the

connected component of X containing $\overset{-1}{f}(W)$. Since $y\in \overline{f(U)}$, it follows from a) and from the definition of the connected component of $y$ that $V =\overline{f(U)}$. Hence V is open and closed in Y.