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Theorem: Let $f: X_1\rightarrow X_2$ be a homeomorphism from $(X_1, \tau_1)$ onto $(X_2, \tau_2)$. $X_1$ is connected if and only if $X_2$ is connected.

Prove this theorem.

My thinking is

  1. If $X_1$ is connected, then $X_2$ is connected.

Proof by contradiction. Since $f$ be a homeomorphism, which means $f$ is bijection and both $f$ and $f^{-1}$ are continuous. Then let $A$, $B$ be a subset of $X_1$. Since $X_1$ is connected, then $A$ and $B$ are not open set such that $A \cap B \neq\emptyset$ and $A \cup B \neq X_1$. There should have $f(A),f(B)$ be a subset of $X_2$, and $f(A)$ and $f(B)$ are not open sets in $X_2$, which is contradiction.

Am I in the right way to solve?

Samuel Adrian Antz
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lulu
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2 Answers2

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This follows from the more general main theorem of connectedness (See for example here.), which can be seen as a generalization of the intermediate value theorem: If $f\colon X_1\rightarrow X_2$ is continuous and $X_1$ is (path)connected, then $f(X_1)\subseteq X_2$ is (path)connected.

If $f$ is a bijection, then you have $f(X_1)=X_2$ and if it is a homeomorphism, then $f^{-1}$ is also continuous and you can apply the theorem to it for the backwards direction.

Proof (for connectedness): Let $U,V\subset f(X_1)$ be open and disjoint sets with $U\cup V=f(X_1)$, then $f^{-1}(U),f^{-1}(V)\subset X_1$ are also open (since $f$ is continuous) and disjoint (as the preimage commutes with intersections: $f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)=f^{-1}(\emptyset)=\emptyset$) and we have (as the preimage also commutes with unions): $$f^{-1}(U)\cup f^{-1}(V) =f^{-1}(U\cup V) =f^{-1}(f(X_1))=X_1.$$ Since $X_1$ is connected, we have $f^{-1}(U)=\emptyset$ (without loss of generality) and therefore $U=\emptyset$. $\square$

The problem with your proof is, that you want to show, that $X_2$ is connected, but you start with subsets of $X_1$ and transfer them to $X_2$ using $f$. The idea is right, but has to be done the exact other way around as not every subset of $X_2$ is the image of a subset of $X_1$. Also, $f$ in general does not commute with intersections and unions (you only get inclusions, which can be proper), while $f^{-1}$ does.

Samuel Adrian Antz
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    Thanks for helping and explaining. Just one things is not too much clear. Why you said $f^{-1}(U) = \emptyset$? I know that since $X_1$ is connected, then $X_1$ only have both open and closed sets are $\emptyset$ and $X_1$. But I am not too much understand why it can has $f^{-1}(U)=\emptyset$ based on the $X_1$ is connected. – lulu Nov 03 '22 at 00:33
  • You're welcome! $f^{-1}(U)$ and $f^{-1})(V)$ are a partion (disjoint union) of $X_1$ into open sets, but since $X_1$ is connected, such a partion mustn't exist and therefore one of the sets has to be empty and the other one has to be $X_1$. I chose $f^{-1}(U)$ for the first and $f^{-1}(V)$ for the second, which is possible without loss of generality as you can just swap the sets with each other. – Samuel Adrian Antz Nov 03 '22 at 00:44
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    OK, thanks so much ! – lulu Nov 03 '22 at 01:38
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We need only to prove: If $f$ is a continuous map from a connected $X$ to $Y$, then $f(X)$ is connected.

Your proof seems a little bit unclear, but using proof by contradiction is good.

Suppose $f(X)$ is not connected, then there exist disjoint nonempty open sets $A$, $B$ in $f(X)$ such that $A\cup B=f(X)$. Then $A=f(X)\cap C$, $B=f(X)\cap D$ for some open sets $C$, $D$ in $Y$. Now $X=f^{-1}(A)\cup f^{-1}(B)=f^{-1}(C)\cup f^{-1}(D)$, which means that $X$ is not connected, a contradiction.

Saunders
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