Null homotopic map from $X$ to $Y$

Question

• Show that a map from $X$ to $Y$ is null homotopic iff it extends to a map from $CX$(cone) to Y.

What does the "extend" mean?

Answer 1

Let's say we define $CX$ as $X\times I/{\sim}$, where $(x, t)\sim (x', t')$ iff $t=t' = 1$. In that case, there is a canonical inclusion $X\hookrightarrow CX$ given by $x\mapsto (x, 0)$.

Now take a map $f:X\to Y$. By the above canonical inclusion we can consider $X\times \{0\}\subseteq CX$ instead of $X$, so $f':X\times \{0\}\to Y$ becomes a map whose domain is a subset of $CX$. We say that $f$ extends to a map on $CX$ if there is a continuous function $g:CX\to Y$ such that $g|_{X\times \{0\}} = f'$. We call it "extend" because that's what we intuitively do: we make the domain of $f$ (or really $f'$) bigger.

Answer 2

In this context, “extend” means that there is a map $$g:CX\rightarrow Y$$ such that $f=g\circ i$, where $i:X\hookrightarrow CX$ is the canonical inclusion.

Intuitively, this is saying that $f$ is nullhomotopic if and only if when the domain of $f$ is identified with the base of the cone, we can “fill in” the values on the full come while remaining continuous.