# Null Homotopy from $S^n$ imply Based Homotopy

Let $g:(S^n, *) \to (X, x)$ be a based map between based topological spaces. Suppose that $g$ is null homotopic in $Top$ (therefore homotopic to a constant map).

I want to prove that in this case $g$ is also null homotopic in $Top_{\bullet}$.Therefore the homotopy map can be choosed such that it conserves the basepoint.

My proof: Let $H: S^n \times I \to X$ homotopy between $g(z) = H(z, 0)$ and constant $c = H(z, 1)$. Since $g$ based we have $H(*,0) = g(*) =x$.

Consider $w:= H \vert {\{*\} \times I}$. Therefore $w(0)=x$.

We define a new homotopy $\widetilde{H}: S^n \times I \to X$ by

$$$(z,t) \to \begin{cases} H(z,2t) & \text{for } 0 \le t \le 1/2 \\ w(2-2t) & \text{for } 1/2 \le t \le 1 \end{cases}$$$

This defines a homotopy from $g$ to constant map to $x$.

But what we really need is a homotopy $\bar{H}: S^n \times I \to X$ with $\bar{H}(*, t) =x$ for all $t \in I$.

By quotient property $\widetilde{H}$ corresponds to a map $m:CS^n \to X$ where $CS^n = S^n \times I /S^n \times \{1\}$. What we really want is that $m$ factorises through $K := S^n \times I /S^n \times \{1\} \cup \{*\} \times I$.

The next step is to call in mind that can inclusion $\{*\} \to S^n$ is a cofibration, therefore also $\{*\} \times I \to S^n \times I$ (see retract characterisation of cofibration).

First question: Is then also $\{*\} \times I \to S^n \times I /S^n \times \{1\} = CS^n$ a cofibration? Why?

If that is the case then one knows that since $\{*\} \times I$ is contractable we have $CS^n = S^n \times I /S^n \times \{1\} \cong S^n \times I /S^n \times \{1\} \cup \{*\} \times I = K$

So we get indeed a map $K \to X$ with $K \cong CS^n$.

Now the second problem: the isomorphism $K \cong CS^n$ isn't canonical so I don't know how extract from $K$ the desired based homotopy $\bar{H}$ between $g$ and $x$.

Your approach is not bad, but is too complicated. Since $g$ is null homotopic in $Top$, it has a continuous extension $G : D^{n+1} \to X$. Define

$$R : S^n \times I \to D^{n+1}, R(y,t) = t \ast + (1-t)y ,$$

$$H = G \circ R : S^n \times I \to X.$$

Then $H(y,0) = G(R(y,0)) = G(y) = g(y)$, $H(y,1) = G(R(y,1)) = G(\ast) = x$, $H(\ast,t) = G(R(\ast,t)) = G(\ast) = x$.

By the way, $\{*\} \times I \to S^n \times I /S^n \times \{1\} = CS^n$ is a cofibration. You can see this directly by the retract characterization (perhaps somewhat tedious) or by observing that $\{*\} \times I$ is mapped to a radial line segment $L$ in $CS^n \approx D^{n+1}$. But you can give $D^{n+1}$ a CW-structure such that $L$ is s a subcomplex, thus $L \hookrightarrow D^{n+1}$ is a cofibration.

• I'm not sure if I understand following statement correctly: "Since $g$ is null homotopic in $Top$, it has a continuous extension $G:D^{n+1} \to X$." Do mean this in the sense that in this case the given homotopy $H: S^n \times I \to X$ factorize throught $D^{n+1} \cong CS^n$ or a possible characterisation of null homotopic maps such that $g$ null homotopic iff it can be extended to $G$ where $G \vert {S^n} = g$. If you mean the second one could you give a reference for a proof of it? only the "$\Rightarrow$" direction – KarlPeter Sep 7 '18 at 0:22
• It is well-known that a map $f : Y \to Z$ is null homotopic if and only if it has continuous extension $F : CY \to Z$, where $Z$ is identified with the base of the cone. Recall $CY = Y \times I / Y \times \{ 1\}$, therefore homotopies $H :Y \times I \to Z$ such that $H_1$ is constant are in 1-1-correspondence with maps $H' : CY \to Z$. Apply this to $Y = S^n$. – Paul Frost Sep 7 '18 at 0:36
• So I mean it in the first sense. – Paul Frost Sep 7 '18 at 0:43