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I am trying to understand the proof of the following theorem: If $f:X \rightarrow Y$ is contiuous and let X be a connected space, then f(X) is a connected space.

In the proof, we construct another function $g:X \rightarrow Z$ obtained by restraigning X. Then, we show that if $Z$ is not connected, then X is not connected.

What I don't understand is why do we construct $g$. My guess is that $f(X) = Z$ but I am not sure about that equality.

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    How are $Z$ and $g$ defined here? – 2017-02-06
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    Yes: they want to show that $f(X)$ is connected, so they have to restrict $f$ to its image and they denote $f(X) = Z$. – 2017-02-06
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    It might be better to use the definition of connected: unable to be written as a disjoint union of nonempty clopen subsets. If $f(X)$ were disconnected, what would that mean for the preimage $f^{-1}(f(X)) = X$? – 2017-02-06
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    @Mnifldz another perfectly okay definition is that $X$ is disconnected iff there exists a continuous onto map $g:X \mapsto \{0,1\}$. Maybe that's what's happening here. – 2017-02-06

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Let $ Y \subset f(X)$ such that Y is both open and closed. Thus by continuity $ f^{-1}(Y)$ is also open and closed in X. But that means it is whole of the space .Thus $Y = f(X)$. Hence proved.

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    The post doesn't ask for a proof, it asks for an explanation of why a proof goes in a particular direction – 2017-02-06