None of the existing questions is exactly answering my question so I'm posting a new question, but feel free to refer me to some already answered question!

In Rudin Theorem 4.22, we know that

If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, and $E$ is a connected subset of $X$, then $f(E)$ is connected.

In the proof, we started with consider $f(E) = A \cup B$, where $A$ and $B$ are nonempty separated subsets. Then put $G = E \cap f^{-1}(A)$ and $H = E \cap f^{-1}(B)$. Then Rudin is claiming that $E = G \cup H$. I'm a little suspicious about this. What if $f$ is non-surjective, then $f^{-1}(A) \cup f^{-1}(B)$ is only a proper subset of $E$? Is there a property of $f$ being continuous that forces $f$ to be 1-1?

  • 1
    $\begingroup$ If $x\in E$, then $f(x)\in f(E)=A\cup B$, so there must be some $a\in A$ or some $b\in B$ such that $f(x)=a$ or $f(x)=b$. But then $x\in f^{-1}(A)$ or $x\in f^{-1}(B)$ showing that $x\in G$ or $x\in H$. $\endgroup$
    – sranthrop
    Dec 13, 2015 at 17:09
  • $\begingroup$ A mapping is not connected, but its image might be $\endgroup$
    – zhw.
    Dec 13, 2015 at 21:02
  • $\begingroup$ Note that no matter which function $f : A \to B$ between any two sets, then we have $$f^{-1}(B)=A$$ This is one of the reasons why pre-images are easier to work with then direct images. $\endgroup$
    – N. S.
    Jul 28, 2017 at 20:16
  • $\begingroup$ Which book are you referring to? $\endgroup$ Oct 16, 2018 at 18:21
  • $\begingroup$ $G$ and $H$ are open sets? $\endgroup$
    – sango
    Jan 19, 2019 at 0:41

4 Answers 4


$f^{-1}(A) \cup f^{-1}(B)$ may be larger than $E$, but it must contain $E$: if $x \in E$, $f(x) \in f(E) = A \cup B$. Then either $f(x) \in A$ or $f(x) \in B$. In the first case, $x \in f^{-1}(A)$, and in the second $x \in f^{-1}(B)$. (The fact that $f^{-1}(A) \cup f^{-1}(B)$ may be larger than $E$ is the reason for intersecting with $E$ when defining $G$ and $H$.)


For all $x\in X$ we have $$ \begin{align} x\in G\cup H &\iff [(x\in E \cap f^{-1}A)\lor (x\in E\cap f^{-1}B] \\ & \iff [x\in E \land (f(x)\in A\lor f(x)\in B)] \\ & \iff [x\in E\land f(x)\in f(E)] \\ & \iff x\in E. \end{align} $$


If $e \in E$, then $f(e) \in f(E) = A \cup B$.
So, $f(e) \in A$ or $f(e) \in B$.
So, $e \in f^{-1}(A)$ or $e \in f^{-1}(B)$.
So, $e \in f^{-1}(A) \cup f^{-1}(B)$.
So, $E \subset f^{-1}(A) \cup f^{-1}(B)$.

So, $E = E \cap (f^{-1}(A) \cup f^{-1}(B)) = (E \cap f^{-1}(A)) \cup (E \cap f^{-1}(B)) = G \cup H$.

$A \cap B = \emptyset$.
So, $f^{-1}(A) \cap f^{-1}(B) = \emptyset$.
So, $\emptyset = E \cap \emptyset = E \cap (f^{-1}(A) \cap f^{-1}(B)) = (E \cap f^{-1}(A)) \cap (E \cap f^{-1}(B)) = G \cap H$


Suppose $E$ is connected subset of $X$. We need to prove that $f(E)$ is connected. Assume the contrary. $f(E)=C\cup D$ forms a separation of $f(E)$. Since $f$ is surjective therefore both $f^{-1}(C)$ and $f^{-1}(D)$ are non-empty subsets of $E$. due to continuity $f^{-1}(C)$ and $f^{-1}(D)$ are open in $X$. $f^{-1}(C) \cap f^{-1}(D)=f^{-1}(C\cap D)=f^{-1}(\phi)=\phi$. Contradicting the fact that $E$ is connected.

  • $\begingroup$ @Zargles I voted to reject your edit as it was not immediately clear that was what the answerer meant. You should instead comment below the post. $\endgroup$
    – user1729
    Nov 5, 2020 at 9:12

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