# proof of image of a connected set is a connected set

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?

• All those sets are within the image, though. Commented Jan 11, 2020 at 13:38

## 2 Answers

$$f^{-1}[A_1] \cap E$$ 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.

• if y is in $f[E]$, why should I conclude that $x=f^{-1}(y)$ is in E), I would say it is in $f^{-1}(f(E))$, which is related to E by $E \subseteq f^{-1}(f(E))$, so that x may fall outside of E, unless they are equal, which happens only if the function is surjective Commented Jan 11, 2020 at 14:11
• @juancarlosvegaoliver $y \in f[E]$ by definition means that there is an $x \in E$ with $y=f(x)$. Recall $f[E] = \{f(x) \mid x \in E\}$. And $E \subseteq f^{-1}[f[E]]$ is always true: $x\in E$ implies $f(x) \in f[E]$ so by definition $x \in f^{-1}[f[E]]$. The other inclusion is related to injectiveness BTW. Commented Jan 11, 2020 at 14:14

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.