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Assume $f(D)\subset U\cup V$ where $U,V$ are open sets in $\Bbb{C}$. $f$ being continuous the inverse image of an open set is an open set and so $f^{-1}(U),\,f^{-1}(V)$ are therefore open sets in $\Bbb{C}$.

Now take $x\in D$, one has $f(x)\in f(D)$ and this means $f(x)\in U$ or $f(x)\in V$ and so $x\in f^{-1}(U)\cup f^{-1}(V)$ and therefore $D\subset f^{-1}(U)\cup f^{-1}(V)$.

But $D$ is connected so it cannot be included in the union of two disjoint open sets and so $\exists a\in f^{-1}(U)\cap f^{-1}(V)$ and thus $f(a)\in U\cap V$. Therefore $U,V$ are not disjoint and $f(D)$ is connected.

From

I am stuck at the third paragraph, could someone explain why it must be that if a connected set can't be included in the union of two disjoint open sets? I get that the definition of connected is that it can't be equal to that but I don't get who it can't even take part!

tryst with freedom
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    You're right: $D \subset f^{-1}(U) \cup f^{-1}(V)$ is not strong enough. Equality should be argued. Fortunately, the other inclusion is trivial, as every element of the inverse image of any set lies in the domain $D$, by definition. – Theo Bendit Jun 12 '22 at 11:48
  • Hmm your comment was very easy for me to understand but I am confused at the first answer posted @TheoBendit are they equivalent? How do they differ – tryst with freedom Jun 12 '22 at 11:58

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Suppose $X$ is a space and $K \subset X$ is a connected subset. That means $K$ is a connected topological space when considered with the subspace topology.

Now suppose we have $K \subset U \cup V$ for $U,V \subset X$ open sets of $X$. Then $U' = U \cap K$ and $V' = V \cap K$ are open subsets sets of $K$ in the subspace topology.

Since $K$ is connected $U'$ and $V'$ cannot be disjoint. It follows $U$ and $V$ cannot be disjoint either, as this would imply $U'$ and $V'$ are disjoint.

Daron
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