A function "extends" to the cone on X

Question

I have the following statement:

A map $f : X \rightarrow Y$ is nullhomotopic if and only if it extends to the cone on $X.$

My problem is that I have no idea what "extends" means in this statement (I understand what the cone is and the definition of nullhomotopic). Someone who can explain this?

Answer 1

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} $$

Since the cone $CX$ over $X$ is the quotient of $X\times I$ by collapsing $X\times\{1\}$ to a point, $X$ embeds as $X\times\{0\}$ in $X\times I$, which then maps homeomorphically to the base of $CX$ under the quotient map $q:X\times I\twoheadrightarrow CX.$ An extension of $f$ to the cone is then a map $h:CX\to Y$ such that the restriction $h|_X$ to this base equals $f$. More precisely, if $i=q\circ(\mathrm{Id}_X,0):X\hookrightarrow CX$ is the embedding, then $h\circ i=f.$

\begin{array}{ccc} X & \ra{f} & Y \\ \da{i} & & \da{=} \\ CX & \ra{h} & Y \end{array}