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Let $ f: X \rightarrow Y $ be a continuous function. And let $ E $ be a connected subset of $ X $.

I want to show that $ f(E) $ is also connected.

I am able to show that if $ X $ is connect if follows that $ f(X) $ is also connect.

I noticed that there are proofs on this site, but they all use a different definition of connected. I want to prove it using the following definition: $ A \subset X $ is connected if there are no non-empty disjoint open sets $ U $ , $ V $ such that $ A = U \cup V $.

My work: For $ f: X \rightarrow Y$ continuous. Suppose $ A \subset X $ is connect. For sake of contradiction assume $ f(A) $ is not connected. Thus there are open sets $ U, V $ such that $ U_1 := U \cap f(A) $ and $ V_1 := V \cap f(A) $, so that $ U_1 $ and $ V_1$ are disjoint and nonempty as well as $ f(A) = U_1 \cup V_1 $.

It follows that $ f^{-1} (U_1) = f^{-1} (U) \cap f^{-1}f(A)) $. However, if $ A \neq X $ I am unable to show that $ f^{-1} (U_1) $ needs to be open or that $ f^{-1} (U_1) \cup f^{-1} (V_1) = f(A) $.

Any help would be much appreciated.

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    $\begingroup$ Does this answer your question? Image of a connected space under a continuous map is connected, proof See also math.stackexchange.com/questions/4555410 $\endgroup$ Sep 18, 2023 at 20:10
  • $\begingroup$ I don't think it does, since this applies to the whole space of $ X $ and not just some connected subset $\endgroup$
    – user007
    Sep 18, 2023 at 20:24
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    $\begingroup$ It does, since when you restrict your continuous map to that connected subset, it remains continuous. $\endgroup$ Sep 18, 2023 at 20:26
  • $\begingroup$ I do not see how you can make sure that the preimage is contained in the restricted set $\endgroup$
    – user007
    Sep 18, 2023 at 20:28
  • $\begingroup$ I don't know which preimage you are talking about, but there is no doubt that the duplicates apply to the restriction $g:A\to Y$ of your map $f:X\to Y.$ Since $A$ is connected and $g$ is continuous, the duplicates prove that $g(A)$ is connected. And $f(A)=g(A).$ $\endgroup$ Sep 18, 2023 at 20:32

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I think you meant "suppose $A\subseteq X$ is connected" (not $A\cup X$).

Hint: instead of focusing on $U_1$, focus on $f^{-1}(U)$. This is open because $f$ is continuous.

Elaboration: Since $U\cup V\supseteq f(A)$, it follows $f^{-1}(U)\cup f^{-1}(V)\supseteq A$ with both terms on the left hand side being open. It remains to show that $f^{-1}(U)\cap A$ and $f^{-1}(V)\cap A$ are disjoint.

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  • $\begingroup$ I see that $ f^{-1} (U) $ is open since $ U $ is open. However, I don't see how this helps me show that $ A $ wouldn't be connected since the only condition $ U $ has to follow is that it is open. $\endgroup$
    – user007
    Sep 18, 2023 at 19:51
  • $\begingroup$ @user007 I elaborated slightly $\endgroup$
    – FShrike
    Sep 18, 2023 at 19:55
  • $\begingroup$ Thanks for the hints. We want to prove $(f^{-1}(U) \cap A) \cap (f^{-1}(V) \cap A) = \emptyset$. That's equivalent to $(f^{-1}(U) \cap f^{-1}(V) \cap A = \emptyset$. Suppose that $ (f^{-1}(U) \cap f^{-1}(V) \cap A \neq \emptyset $. We can follow that $U \cap V \cap f(A) \neq \emptyset$. We also have $U \cap V_1 \neq \emptyset$. But this implies that there's an element $a$ such that $ a \in V_1$ thus in $ f(A)$ and $ V $ but also in $U$. Therefore it is in $U_1$. But that is a contradiction. Therefore, $(f^{-1}(U) \cap A) \cap (f^{-1}(V) \cap A) = \emptyset$. But that is a contradiction. $\endgroup$
    – user007
    Sep 18, 2023 at 20:18
  • $\begingroup$ Is that correct? $\endgroup$
    – user007
    Sep 18, 2023 at 20:18
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    $\begingroup$ @user007 That looks good. I might just say, suppose $f^{-1}(U)\cap f^{-1}(V)\cap A$ is nonempty. Then there is $x$ in all of these. $f(x)\in U$ and $f(x)\in V$ and $f(x)\in f(A)$ so $f(x)\in(U\cap f(A))\cap(V\cap f(A))$ but by assumption these are disjoint $\endgroup$
    – FShrike
    Sep 18, 2023 at 20:42

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