A contractible space is path connected.

Question

Let the space be $X$ and $\rm{id} \simeq x_0$ where $\rm{id}$ is the identity map on $X$ and $\rm x_0$ is some fixed point in $X$. How do I show that for any two points $a,b \in X$ there is a continuous path $f:[0,1] \to X$ such that $f(0) = a, f(1) = b$?

Answer 1

Consider a homotopy between the identity map on $X$ and the point $x_0$. Choose a point $a$ and using this homotopy find a path between $a$ and $x_0$.

Answer 2

If $X$ is contractible, say to $x_0\in X$, there is a continuous function $$ f: I\times X\to X, $$ where $I=[0,1]$, such that $f(0,x)=x$ and $f(1,x)=x_0$, for all $x\in X$.

Now let $a,b\in X$, and define the path $\gamma: I\to X$ as $$ \gamma(t)=\left\{\begin{array}{lll} f(2t,a) & \text{if} & t\in[0,1/2], \\ f(2-2t,b) & \text{if} & t\in[1/2,1]. \end{array}\right. $$ Clearly $\gamma$ is continuous (Pasting Lemma), and also $\gamma(0)=a$ and $\gamma(1)=b$.

Answer 3

Take any two points $p$ and $q$ and a homotopy $F(x,t)$ such that $F(x,0)=x$ and $F(x,1)=z$ for some fixed point $z$ . Then $F(p,t)$ as a function in $t$ defines a path from $p$ to $z$ , and $F(q,t)$ defines a path from $q$ to $z$ , so that all three points $p$ , $q$ and $z$ are in the same path connected component.