The cone of a topological space is always path-connected I'm having trouble proving that if $X$ is a topological space, then $C(X)$ (defined as $X \times [0,1] /\sim$ where $(x_1,t_1)\sim (x_2,t_2)$ iff they are equal or $t_1=0=t_2$) is path-connected.
I understand the definition and I know I must take two different points in the cone and construct a function from $[0,1]$ to $C(X)$. The problem is that because $C(X)$ is a quotient, then I don't quite see how points look like and how to join them with a path.
Thank you.
 A: First note that it suffices to show that there exists a point $p$ to which any point in $CX$ can be connected by a continuous path.  (Make sure you understand why this is true!)  In the cone, we can take this "connecting point" to be the vertex of the cone, i.e., the point $p = [x,0]$ (where $x$ is any point in $X$). Can any point in $CX$ be connected to the vertex $p$ by means of a continuous path?  Definitely.
To see this, let $[x,t]$ be a point in $CX$ and consider the map $\lambda : I \rightarrow X \times I$ defined by $\lambda(s) = (x,(1-s)t)$. Since both coordinates of $\lambda$ are continuous, $\lambda$ itself is continuous (i.e., $\lambda$ is a continuous path in $X \times I$ that joins the point $(x,t)$ to the point $(x,0)$ (this is still true if $t = 0$.)  Let $\pi : X \times I \rightarrow CX$ be the quotient map.  Since $\pi$ is continuous, the composition $\pi \circ \lambda$ is also continuous.  This composition is thus a continuous path that joins $[x,t]$ to $p = [x,0]$.
Since we've shown than an arbitrary point $[x,t]$ in $CX$ can be joined to $p$, we can conclude that $CX$ is indeed path-connected.
