---- > [!definition] Definition. ([[map on path components induced by continuous function]]) > Let $X,Y$ be [[topological space|topological spaces]] and $h:X \to Y$ [[continuous]]. Then there is a [[well-defined]] function $\begin{align} \pi_{0}(h) : \pi_{0}(X) \to \pi_{0}(Y) \\ [x] \mapsto [h(x)], \end{align}$ > called the **map on $\pi_{0}$ induced by $h$**. ^definition > [!justification] > We must show the proposed function is [[well-defined]], so let $[x]=[x'] \in \pi_{0}(X)$. This means there is a [[parameterized curve]] $\gamma:I \to X$ from $x$ to $x'$. So $h \circ \gamma$ is a [[parameterized curve]] from $h(x)$ to $h(x')$ and therefore $[h(x)]=[h(x')]$ as required. ^justification > [!basicproperties] Functorial properties of the induced map. > 1. If $h:X \to Y$ and $k:Y \to Z$ are [[continuous]], then $\pi_{0}(k \circ h)=\pi_{0}(k) \circ\pi_{0}(h)$; > 2. The [[identity map]] $\id_{X}$ induces $\id_{\pi_{0}(X)}$. > [!proof] Proof of functorial properties. > >This is trivial. >1. Let $h,k$ be [[continuous]]. Then $\begin{align} \pi_{0}(k \circ h)([x]) = & [k\big([h(x)]\big)] \\ = & [k(\pi_{0}(h)(x))] \\ = & \pi_{0}(k) \circ \pi_{0}(h)(x) \end{align}$ > 2. Clearly $\pi_{0}(\id_{X})(x)=[\id_{X}(x)]=[x]$. > [!basicproperties] Other Properties. > - If $f:X \to Y$ is [[homotopy|homotopic]] to some $g$, then $\pi_{0}(f)=\pi_{0}(g)$ > - If $f: X \to Y$ is a [[homotopy equivalent|homotopy equivalence]] then $\pi_{0}(f):\pi_{0}(X) \to \pi_{0}(Y)$ is a [[bijection]]. ^properties > [!proof] Proof of other properties. > - Fix $x \in X$ and suppose $f \simeq g$ via $H$. Define a [[parameterized curve]] $\gamma:I \to Y$ by $\gamma(t):=H(x,t)$. Since $\gamma(0)=H(x, 0)=f(x)$ and $\gamma(1)=H(x,1)=g(x)$, $\gamma$ is a [[parameterized curve]] from $f(x)$ to $g(x)$. Thus $\pi_{0}(f)([x])=[f(x)]=[g(x)]=\pi_{0}(g)([x])$. > - Let $g:Y \to X$ be a [[homotopy equivalent|homotopy inverse]] of $f$. Then using the above result we have $\pi_{0}(f) \circ \pi_{0}(g) = \pi_{0}(f \circ g)=\pi_{0}(\id_{Y} )=\id_{\pi_{0}(Y)}$ and likewise$\pi_{0}(g) \circ \pi_{0}(f)=\pi_{0}(g \circ f)=\pi_{0}(\id_{X})=\id_{\pi_{0}(X)}$ so $\pi_{0}(g)$ is the inverse to $\pi_{0}(f)$. ^proof ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #reformatrevisebatch03