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Here is the question:

Show that if $f: X \rightarrow Y$ is any continuous function, and if $X$ is connected, then its image $f(X) \subseteq Y$ (which has the subspace topology) is also connected.

Where we have the following:

we say a space $X$ is connected if the only separations that $X$ has are trivial separations.

A separation of a space $X$ is a continuous function $f: X \rightarrow Z,$ where $Z$ is the discrete two-point space. we say that a separation $f: X \rightarrow Z,$ of the space $X$ is a trivial separation if it is a constant function.

Still I do not know how to prove the question, could anyone help me please?

Emptymind
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    You can check out this question: https://math.stackexchange.com/questions/1573795/proof-of-the-continuous-image-of-a-connected-set-is-connected It's posed in metric spaces, but proven for general topological spaces. – user754697 Mar 03 '20 at 04:51

2 Answers2

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Suppose $X$ is connected, $f:X\to Y$ is continuous, and $g$ is a continuous function from the image of $f$ to the discrete two-point space $Z$. Then $g\circ f:X\to Z$ is a composition of continuous maps, so continuous, and $X$ is connected, so the composition is constant, so $g$ is constant.

Gerry Myerson
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  • I do not understand why the composition being constant leads to that $g$ is constant ... could you explain this more please? – Emptymind Mar 03 '20 at 13:46
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    Suppose $g$ is not constant. Then there exist $a,b$ in the image of $f$ such that $g(a)\ne g(b)$. Since $a,b$ are in the image of $f$, there exist $c,d$ in $X$ such that $f(c)=a$ and $f(d)=b$. Then $(g\circ f)(c)=g(a)\ne g(b)=(g\circ f)(d)$, so $g\circ f$ is not constant, contradiction. – Gerry Myerson Mar 03 '20 at 21:30
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If you can disconnect the image, $f(C)=U\cup V$, with $U$ and $V$ open and disjoint, then $f^{-1}(U)$ and $f^{-1}(V)$ disconnect $C$.

The existence of a separation of $X$ is equivalent to that of a pair of disjoint open sets whose union is $X$.