# Proving that the image of a connected set is connected

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.

• Does this answer your question? Image of a connected space under a continuous map is connected, proof See also math.stackexchange.com/questions/4555410 Sep 18, 2023 at 20:10
• I don't think it does, since this applies to the whole space of $X$ and not just some connected subset Sep 18, 2023 at 20:24
• It does, since when you restrict your continuous map to that connected subset, it remains continuous. Sep 18, 2023 at 20:26
• I do not see how you can make sure that the preimage is contained in the restricted set Sep 18, 2023 at 20:28
• 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 $f(A)=g(A).$ Sep 18, 2023 at 20:32

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.

• 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. Sep 18, 2023 at 19:51
• @user007 I elaborated slightly Sep 18, 2023 at 19:55
• 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. Sep 18, 2023 at 20:18
• Is that correct? Sep 18, 2023 at 20:18
• @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 Sep 18, 2023 at 20:42