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I'm reading the following theorem:

I don't understand what would go wrong if you would leave out the part that I marked green. I would think you could safely leave it out and replace the second to last sentence with this sentence:

"Then $f^{-1}(A)$ and $f^{-1}(B)$ are disjoint sets, whose union is $X$; they are open in $X$ because $f$ is continuous, and nonempty because $A\cup B=Z=f(X)$."

Or am I missing something ?

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up vote 3 down vote accepted

In this case, because of the very strong connection between the continuous function $f$ and the subspace $Z$ of $Y$ under consideration, your change would work, with a bit of extra argumentation.

Recall that the definition of disconnectedness of a subset $Z$ of a space $Y$ is:

there are open subsets $U , V$ of $Y$ such that

  • $Z \subseteq U \cup V$; and
  • $U \cap Z \neq \varnothing \neq V \cap Z$
  • $(U \cap V ) \cap Z \neq \varnothing$.

Of course, this is equivalent to saying that $Z$ is disconnected (in the subspace topology).

You would now consider $f^{-1} [ U ]$ and $f^{-1} [ V ]$ and conclude that these are disjoint open subsets of $X$ whose union is $X$. That they are open follows from continuity of $f$. That their union is $X$ follows from the fact that $U \cup V \supseteq f [ X ]$. But to conclude that $\varnothing = f^{-1} [ U ] \cap f^{-1} [ V ] = f^{-1} [ U \cap V ]$ we have to make a bit of extra argumentation to conclude that this is empty (since it is not necessarily true that $U \cap V = \varnothing$. That extra bit is something to the effect of

Clearly $f^{-1} [ U \cap V ] = f^{-1} [ ( U \cap V ) \cap f[X] ]$ and since $Z = f[X]$ and $( U \cap V ) \cap Z = \varnothing$ it follows that $f^{-1} [ U \cap V ] = f^{-1} [ \varnothing ] = \varnothing$.

A slight technicality that adds a bit of unnecessary complexity to the proof, in my opinion.

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It's just convenience. If you consider the subspace $f(X)$ embedded in $Y$, you can't simply say "Suppose $f(X) = A\cup B$ is a separation of $f(X)$ into two disjoint nonempty open sets", since $f(X)$ generally is not open in $Y$. You need to say "relatively open", and then remark that the preimage of a relatively open subset of $f(X)$ is open in $X$, or take a covering of $f(X)$ by open sets, $f(X) \subset A\cup B$, where $A,B$ are open sets in $Y$ with $A\cap B \cap f(X) = \varnothing$ and $A\cap f(X) \neq \varnothing \neq B\cap f(X)$. There's not much difference to considering the induced surjective map to $Z=f(X)$ either way.

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