The Stacks project

Proof. Let $A \subset f(E)$ an open and closed subset of $f(E)$. Because $f$ is continuous, $f^{-1}(A)$ is an open and closed subset of $E$. As $E$ is connected, $f^{-1}(A) = \emptyset$ or $f^{-1}(A) = E$. However, $A\subset f(E)$ implies that $A = f(f^{-1}(A))$. Indeed, if $x\in f(f^{-1}(A))$ then there is $y\in f^{-1}(A)$ such that $f(y) = x$ and because $y\in f^{-1}(A)$, we have $f(y) \in A$ i.e. $x\in A$. Reciprocally, if $x\in A$, $A\subset f(E)$ implies that there is $y\in E$ such that $f(y) = x$. Therefore $y\in f^{-1}(A)$, and then $x\in f(f^{-1}(A))$. Thus $A = \emptyset$ or $A = f(E)$ proving that $f(E)$ is connected. $\square$

Proof. Let $\overline{T}$ be the closure of the connected subset $T$. Suppose $\overline{T} = T_1 \amalg T_2$ with $T_ i \subset \overline{T}$ open and closed. Then $T = (T\cap T_1) \amalg (T \cap T_2)$. Hence $T$ equals one of the two, say $T = T_1 \cap T$. Thus $\overline{T} \subset T_1$. This implies (1) and (2).

Let $A$ be a nonempty set of connected subsets of $X$ such that $\Omega = \bigcap _{T \in A} T$ is nonempty. We claim $E = \bigcup _{T \in A} T$ is connected. Namely, $E$ is nonempty as it contains $\Omega$. Say $E = E_1 \amalg E_2$ with $E_ i$ closed in $E$. We may assume $E_1$ meets $\Omega$ (after renumbering). Then each $T \in A$ meets $E_1$ and hence must be contained in $E_1$ as $T$ is connected. Hence $E \subset E_1$ which proves the claim.

Let $W \subset X$ be a nonempty connected subset. If we apply the result of the previous paragraph to the set of all connected subsets of $X$ containing $W$, then we see that $E$ is a connected component of $X$. This implies existence and uniqueness in (3).

Let $x \in X$. Taking $W = \{ x\}$ in the previous paragraph we see that $x$ is contained in a unique connected component of $X$. Any two distinct connected components must be disjoint (by the result of the second paragraph).

To get an example where connected components are not open, just take an infinite product $\prod _{n \in \mathbf{N}} \{ 0, 1\}$ with the product topology. Its connected components are singletons, which are not open. $\square$

Proof. Let $T \subset Y$ be a connected component. Note that $T$ is closed, see Lemma 5.7.3. The lemma follows if we show that $f^{-1}(T)$ is connected because any connected subset of $X$ maps into a connected component of $Y$ by Lemma 5.7.2. Suppose that $f^{-1}(T) = Z_1 \amalg Z_2$ with $Z_1$, $Z_2$ closed. For any $t \in T$ we see that $f^{-1}(\{ t\} ) = Z_1 \cap f^{-1}(\{ t\} ) \amalg Z_2 \cap f^{-1}(\{ t\} )$. By (1) we see $f^{-1}(\{ t\} )$ is connected we conclude that either $f^{-1}(\{ t\} ) \subset Z_1$ or $f^{-1}(\{ t\} ) \subset Z_2$. In other words $T = T_1 \amalg T_2$ with $f^{-1}(T_ i) = Z_ i$. By (2) we conclude that $T_ i$ is closed in $Y$. Hence either $T_1 = \emptyset$ or $T_2 = \emptyset$ as desired. $\square$

Proof. This is a special case of Lemma 5.7.5. $\square$

Proof. If the topological space $X$ has at least $N$ connected components for some $N \in \mathbf{N}$, we find by induction a decomposition $X = X_1 \amalg \ldots \amalg X_ N$ of $X$ as a disjoint union of $N$ nonempty open and closed subsets $X_1, \ldots , X_ N$ of $X$. As $f$ is open and closed, each $f(X_ i)$ is a nonempty open and closed subset of $Y$ and is hence equal to $Y$. In particular the intersection $X_ i \cap f^{-1}(y)$ is nonempty for each $1 \leq i \leq N$. Hence $f^{-1}(y)$ has at least $N$ elements. $\square$

Proof. By Lemma 5.7.5 the connected components of $\pi _0(X)$ are the singletons. We omit the proof of the second statement. $\square$

Proof. For all $x\in X$ let write $\mathcal{N}(x)$ the fundamental system of connected neighbourhoods of $x$ and let $U\subset X$ be an open subset of $X$. Then for all $x\in U$, $U$ is a neighbourhood of $x$, so the set $\{ V \in \mathcal{N}(x) | V\subset U\}$ is not empty and is a fundamental system of connected neighbourhoods of $x$ in $U$. Thus $U$ is locally connected and it proves (1).

Let $x \in \mathcal{C} \subset X$ where $\mathcal{C}$ is the connected component of $x$. Because $X$ is locally connected, there exists $\mathcal{N}$ a connected neighbourhood of $x$. Therefore by the definition of a connected component, we have $\mathcal{N} \subset \mathcal{C}$ and then $\mathcal{C}$ is a neighbourhood of $x$. It implies that $\mathcal{C}$ is a neighbourhood of each of his point, in other words $\mathcal{C}$ is open and (2) is proven. $\square$