When can homotopy be upgraded to a pointed homotopy?

Question

Suppose that $$f,g: (X,x_0) \rightrightarrows (Y,y_0)$$ are freely homotopic pointed maps via a homotopy $H : f \rightarrow g$. Furthermore, suppose that $H(x_0,t)$ is a nullhomotopic loop based in $y_0$. What I was curious is weather this enough for $f$ to be pointed homotopic to $g$.

I was messing around with some lift diagrams and couldn't found a decent one. Any help is appreciated.

Answer 1

The answer is no, it is not enough for $H(x_0, t)$ to be nullhomotopic.

Question 6a in Chapter 0 of Hatcher's Algebraic Topology book goes as follows:

6. (a) Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the horizontal segment $[0, 1]\times \{0\}$ together with all the vertical segments $\{r\}\times [0, 1 − r ]$ for $r$ a rational number in $[0, 1]$ . Show that $X$ deformation retracts to any point in the segment $[0, 1]\times \{0\}$ , but not to any other point. [See the preceding problem.]

Take $X$ as in the question, define $f$ to be the identity map on $X%$, and $g$ to be a constant map to a point $y$ outside of the segment $[0, 1]\times \{0\}$. The second part of 6(a) is equivalent to saying that there is no pointed homotopy from $f$ to $g$.

Nevertheless, $f$ and $g$ are homotopic with $H(y, t)$ nullhomotopic. To see this, use the first part of 6(a) to construct a deformation retract to a point $x$ on the segment $[0, 1]\times \{0\}$, and compose it with a homotopy that takes $x$ to $y$.