A simple intuitive example of a nullhomotopic map

algebraic-topologyhomotopy-theory

I'm just in the first week of an Algebraic Topology class and have been presented with the following definition.

A map is nullhomotopic if it is homotopic to a constant map. I.e. $f: X \to Y$ is nullhomotopic if, and only if, there exists a $y_0 \in Y$, $F: X \times [0,1] \to Y$ with $F(x,0) = f(x)$ and $F(x,1) = y_0$.

From looking around on MSE I can see similar questions have been asked but answers often relate to concepts I have not yet seen (e.g. fundamental group). If possible, I am just looking for a simple example of functions $f$ and $F$ to help build a basic (if possible visual) intuition of what types of functions and spaces would satisfy this definition. I'm happy with the definition of homotopic, but I'm finding it hard to visualise nullhomotopic.

I have been presented with the fact that the identity on $\mathbb{R}^n$ is an example of a null homotopic map, but even this simple example I am struggling to explicitly match with the provided definition.

Thanks in advance.

Best Answer

I think the best way to approach this is to first think about null-homotopic loops, which in my opinion have a much cleaner visual intuition than more general functions.

Let's consider a continuous map $f:\mathbb{S}^1\rightarrow \mathbb{R}^2$. For sake of simplicity let's assume $f$ is an embedding, so $f(\mathbb{S}^1)\simeq \mathbb{S}^1$. Visually, all this means is we have placed a circle in $\mathbb{R}^2$. Now, pick some point that the loop passes through, and imagine a someone standing at this point holding onto the loop. Imagine the loop is made of rope. The person begins pulling on both sides of the loop, and we can see that they are eventually able to pull in the entirety of the rope so that it's in a little pile at their feet. This loop will be null-homotopic: the person can "pull on the loop" enough so that eventually all the rope is at the location they are standing at. Note that it doesn't matter "how quickly" the person pulls the rope or if they "wiggle" the rope as they pull it in, all that matters is they can eventually pull all the rope into their location. Since $f$ determines this loop, we would say that "$f$ is null-homotopic." The map describing how the person pulls on the loop would be the homotopy such that $H(x,0)=f(x)$ and $H(x,1)=y$, where $y$ is the point the person is standing at. Note that although in the illustration the person stays in the same spot, all that matters is their end spot; they could move around while pulling on the loop.

Now imagine a large pull being placed in the center of the loop. Then no matter how long the person pulls on the loop/rope and even if they change their position, the rope will always get stuck around the pole eventually and thus the person can never pull the loop completely into their location. This means the loop is not null-homotopic.

Now, given a continuous map $g:X\rightarrow Y$ between two topological spaces, the intuition of "null-homotopic" is similar, although not quite as easy to visualize. Namely, I tend to think of someone "pulling on" the image of $g$, and the map is null-homotopic iff the person can pull all of $g$, i.e. all of the image of $g$ into their lap.

I present some mathematical examples below of null-homotopic maps. In considering these, use the illustration I presented above.

Let $V$ be any (connected) topological vector space (a vector space equipped with a topology such that addition and scalar multiplication are continuous), and let $f:X\rightarrow V$ be any continuous map. Then define the homotopy $H:X\times [0,1]\rightarrow V$ by $H(x,t)=(1-t)f(x)+t y$, where $y \in V$ is arbitrary.

Let $f:X\rightarrow\mathbb{S}^n$ be such that $f(X) \subsetneq \mathbb{S}^n$. Then there is some point $x \in \mathbb{S}^n$ such that $f^{-1} (x)=\varnothing$. Let $\phi:\mathbb{S}^n\setminus \{x\}\rightarrow\mathbb{R}^n$ be the stereographic projection about this point. Let $y\in \mathbb{S}^n$ be such that $y\neq x$. Then we have a homotopy $H(x,t)=\phi^{-1}((1-t)(\phi\circ f)(x)+t\phi(y))$, from which we can see that $f$ is null-homotopic.

EDIT: I do need to point out that “pulling on the image” is not entirely what happens, as you can have maps with the same image that are not homotopic. Again, the visual intuition I gave is more for a rough idea of what’s happening.

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