📂Topology

A Surjective Continuous Function Preserves Connectedness

Theorem

For a connected space $X$, if $f : X \to Y$ is a surjective continuous function, then $Y$ is a connected space.

Explanation

Since similar-sounding words like connected and continuous are mixed together, it can be a little confusing. Usually memorizing them in English solves this, but the English words used in this theorem are Connected and Continuous, so it isn’t much help.

Proof

Assume that $Y$ is not a connected space. Then there exist open proper subsets $A,B \subset Y$ satisfying $$ A \cap B = \emptyset \\ A \cup B = Y $$ Since $f$ is a surjective function, $f^{-1}(A)$ and $f^{-1}(B)$ are not empty.

If $f$ is a continuous function, then for every open set $V \subset Y$, $f^{-1} (V)$ is an open set in $X$.

Since $f$ is a continuous function, $f^{-1}(A)$ and $f^{-1}(B)$ are open sets in $X$. However, taking their intersection gives $$ f^{-1} (A) \cap f^{-1} (B) = f^{-1} (A \cap B) = f^{-1} ( \emptyset ) = \emptyset $$ while taking their union gives $$ f^{-1} (A) \cup f^{-1} (B) = f^{-1} (A \cup B) = f^{-1} ( Y ) = X $$ Therefore $X$ is a disconnected space, which is a contradiction.

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As you can tell by examining the proof carefully, it is exactly the same as the method for showing that connectedness is a topological property. Rather than the theorem as a whole, the corollary below, which takes on a stronger character as a fact, will be more convenient to use.

Corollary

For a connected space $X$, if $f : X \to Y$ is a continuous function, then $f(X)$ is a connected space¹.