Preservation of Connectivity under Continuous Maps

One particularly nice property of connected topological spaces $X$ is that if $f : X \to Y$ is a continuous map then the range $f(X)$ is also connected. We prove this result below.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$ be a continuous map. If $X$ is connected then $f(X)$ is connected.

• Proof: Let $X$ be connected and suppose that $f(X)$ is instead disconnected. We will show that a contradiction arises. Let $\{ A, B \}$ be a separation of $f(X)$, i.e., $A, B \subset f(X)$ are open in $f(X)$ with the subspace topology, $A, B \neq \emptyset$, $A \cap B = \emptyset$, and:

(1)

\begin{align} \quad f(X) = A \cup B \end{align}

• Since $f$ is continuous, we see that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X$. We claim that $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X$.

• We have already established that $f^{-1}(A)$ and $f^{-1}(B)$ are open in $X$. Furthermore, $f^{-1}(A) \neq \emptyset$ and $f^{-1}(B) \neq \emptyset$. Clearly since $f(X) = A \cup B$ we see that $X = f^{-1}(A) \cup f^{-1}(B)$. All we need to check is that $f^{-1}(A) \cap f^{-1}(B) = \emptyset$.

• Suppose not. Then there exists an $x \in f^{-1}(A) \cap f^{-1}(B)$, so $f(x) \in A$ and $f(x) \in B$. But this implies that $A \cap B \neq \emptyset$ which is a contradiction. Thus, $\{ f^{-1}(A), f^{-1}(B) \}$ is a separation of $X$ which shows that $X$ is disconnected. But this is a contradiction.

• Therefore the assumption that $f(X)$ is disconnected is false. So if $X$ is a connected topological space and $f : X \to Y$ is continuous then $f(X)$ is also connected. $\blacksquare$

Corollary 1: If $X$ and $Y$ are topological spaces and $f : X \to Y$ is a homeomorphism between $X$ and $Y$ then if $X$ is connected then $Y$ is connected.

• Proof: If $f : X \to Y$ is a homeomorphism between $X$ and $Y$ then by definition $f$ is a continuous function. Furthermore, since $f$ is bijective, $f(X) = Y$. By Theorem 1 we immediately get that $f(X) = Y$ is connected. $\blacksquare$