proof of image of a connected set is a connected set

Question

For the following theorem

Let $f $ be a continuous function between topological spaces $f: (X, \tau)\rightarrow(Y,\tau')$ and $E$ a connected set, then $f(E)$ is a connected set.

this is the proof given in my lecture notes:

Let's suppose by absurd that f(E) is disconnected. Then, there exist $A_1, A_2 \in τ'$, such that $A_1 \cap f(E)$ and $A_2 \cap f(E)$ disconnect $f(E)$, ie.

$(A_1 \cap f(E)) \cap$ $(A_2 \cap f(E))$ = $A_1 \cap A_2 \cap f(E)=\emptyset$...($\alpha$)
and

$(A_1 \cap f(E)) \cup$ $(A_2 \cap f(E))=f(E)$. ...($\beta$)

The last statement implies $f(E) ⊆ A_1 \cup A_2$...($\gamma$)

Then $f^{−1}(A_1)$ and $f^{−1} (A_2)$ are open sets in $\tau$, because f is continuous. I must prove that the open sets in the induced topology over E: $f^{−1}(A_1) \cap E$ and $f^{−1}(A_2) \cap E$ disconnect E, that is that they are non-empty, disjoint and their union is $E$

From $(\alpha )$ ,

it follows that $f^{−1}(A_1) \cap f^{−1}(A_2)\cap E =\emptyset $ , then $(f^{−1}(A_1)\cap E) \cap (f^{−1}(A_2)\cap E) =\emptyset $, so they are disjoint.

and from $(\gamma) $

$E ⊆ f^{−1}f(E) ⊆ f^{−1}(A_1 \cup A_2) = f^{−1}(A_1) \cup f^{−1}(A_2)$

then $E=E \cap (f^{−1}(A_1) \cup f^{−1}(A_2))= (E \cap f^{−1}(A_1)) \cup (E \cap f^{−1}(A_2)))$ , so their union is $E$.

To complete the proof and get a contradiction (because $E$ is connected by hypothesis) ,one thing is missing. I can't figure out how to justify $(E \cap f^{−1}(A_1))$ and $(E \cap f^{−1}(A_2))$ are non-empty.

My professor said it is because since $A_1 \cap f(E)$ and $A_2 \cap f(E)$ are non-empty in $\tau'$, then the preimages should be non-empty in $\tau$ but I am suspicious of this, because the preimage of a non-empty set can be the empty set, if the function is not surjective and taking the preimage ,only yields an inclusion:

$ f^{−1}(A_1 \cap f(E))= f^{−1}(A_1) \cap (f^{−1}f(E))$ and $f^{−1}(A_1) \cap E ⊆ f^{−1}(A_1) \cap (f^{−1}f(E))$

Any idea?

Answer 1

$f^{-1}[A_1] \cap E$ is non-empty: we know that $f[E] \cap A_1$ is non-empty, say that $y$ is in this intersection, so that we can write it as $y=f(x)$ for some $x \in E$ (as $y$ is in $f[E]$) and $y \in A_1$. Now $x \in f^{-1}[A_1]$ as its image $f(x)=y \in A_1$ and so $x \in f^{-1}[A_1] \cap E$ which is thus non-empty.

The same argument holds for $E \cap f^{-1}[A_2]$ of course. You don't need to justify every step by general set formulae (though we could), sometimes just straightforward reasoning will do it.

Another type of solution: a space $X$ is connected iff every continuous $f: X \to D(2)$ is constant (where $D(2)$ is a two point set in the discrete topology, say $\{0,1\}$). This is a well-known alternative characterisation of connectedness. Now if $E$ is connected, let $g: f[E] \to D(2)$ be continuous. Then $g \circ f: E \to D(2)$ also is continuous (composition) and as $E$ is connected $g \circ f$ is constant. It follows that $g$ is constant too. Hence $f[E]$ is connected.

Answer 2

The thing is that the 'restriction' of $f$ to $f(E)$ $$f\colon E\rightarrow f(E)$$ is indeed surjective, so the problem that you pointed out does not occur.