Continuous Functions Preserve Connectedness

Last updated Nov 1, 2022

Let $X, Y$ be Topological Spaces and suppose $X$ be Connected. Then if $f: X \to Y$ is a Continuous Function, $f(X)$ is also Connected.

We’ll prove by Contraposition. We assume $f(X)$ is not Connected. That is there exist $U, V \subset Y$ Open so that $U \cap f(X) \sqcup V \cap f(X) = f(X)$ and $U \cap f(X),V \cap f(X)$ not $\emptyset, f(X)$. Then, because $f$ is Continuous Function, $$f^{-1}(U \cap f(X)) = f^{-1}(U) \cap X = f^{-1}(U)$$ is Open. Likewise for $V \cap f(X)$. Let $$\begin{align*} U’ := f^{-1}(U) (= f^{-1}(U \cap f(X)))\\ V’ := f^{-1}(V) (= f^{-1}(V \cap f(X))) \end{align*}$$ Since $U \cap f(X)$ is disjoint from $V \cap f(X)$, we have that $f^{-1}(U) \cap f^{-1}(V) = \emptyset$. But since $U \cup V \supset f(X)$, $f^{-1}(U) \cup f^{-1}(V) \supset X$, and is thus equal to $X$. Since $U \cap f(X)$ neither Empty Set nor $f(X)$, we have that both $f^{-1}(U)$ and $f^{-1}(V)$ are Nonempty. Thus $X$ is not Connected. $\blacksquare$