Explain "homotopy" to me [closed]

I have been struggling with general topology and now, algebraic topology is simply murder. Some people seem to get on alright, but I am not one of them unfortunately.

Please, the answer I need is ideally something very elaborate and extensive, possibly with easy-to-understand examples. Rewriting the definitions with concrete mathematical language and symbols won't help(those are readily accessible in my lecture notes). I want an explanation of what is happening and which part of the solid definitions are telling me so.

One thing I would like to make clear is, I know the definitions, but I don't understand them; I can re-iterate them upon request, consult my lecture notes. Issue is I am writing down something I don't know what it means. It's like writing ancient Greek. I can remember the shapes of each character, their order and write some sentence down. But that's it. Can I explain it to someone with my own words? Break it down? Absolutely not. That is why I am here to ask someone who fully understands these ideas to do exactly that for me: break it down. Go in slow motion. Show me the moves and codes behind it.

Here's the definition I have for homotopy

A homotopy between maps $f,g:X \rightarrow Y$ is a map $h: X \times I \rightarrow Y$ such that

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ where $x \in X$ and $I=[0,1]$. We say maps $f,g$ are homotopic.

It adds that,

A homotopy deforms the map $f$ continuously to $g$.

So, I am given two sets $X,Y$ whatever they are and two maps $f,g$ that takes an element of $X$ to an element of $Y$. I don't know if $f,g$ are bijective, only injective or surjective or whatever. No information on that. Just maps.

And this "homotopy" is a ... "map between maps"? And even if so, what exactly is it doing? All the explanation seems to be done with this single line

$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$

but no, I don't get what's happening. So $X$ has a bunch of elements $x$, and the product space $X \times I$ gives me elements of the form $\{x,t\} \in X \times I$. Okay. But this map $h$ qualifies as a homotopy as long as the above holds? Then why not just always define $h$ as $h(x,t)=f(x)$ for any $t \neq 1$ and $h(x,t)=g(x)$ for $t=1$, just as how we might define a piecewise function? Then I can define this "homotopy" on any maps.

Sure, it adds "deforms $f$ CONTINUOUSLY to $g$" but how is that stated in the definition itself? I see it nowhere.

Here is an example in my notes which didn't help me understand the definition,

Take $X=\{x\}$ the space with a single element $x$. Then a map $f:X \rightarrow Y$ is the same as and element $f(x) \in Y$. A homotopy $h: f \simeq g: X \rightarrow Y $ is the same as a path $h: I \rightarrow Y$ with initial point $h(0)=f(x)$ and terminal point $h(1)=g(x) \in Y$. A homotopy $h: f \simeq f: X \rightarrow Y$ is the same as a closed path $h: I \rightarrow Y$.

Well, first off what is a "path"? Intuition also doesn't make sense because when mapping one element to another, how can there be different "paths"? $x$ goes to $y$. Done. It's not like going from England to Singapore via either Amsterdam or Frankfurt (thus different paths) is it? Unless it's a map $X \rightarrow Z \rightarrow Y$ and telling me that $x \in X$ goes to $z_1 \in Z$ then to $y \in Y$ or $z_2 \in Z$ and then to $y \in Y$, that might be different paths from $x$ to $y$. But here, it is talking about $X$ and $Y$ only.

And why is this ignoring $x$? It says $h: I \rightarrow Y$? The problem I also have is, this is labelled "example" but it's not specific at all. "$h(0)=f(x)$ and $h(1)=g(x)$" over. So? What is this $h$ and how has it been defined? As a map, as a function of some sort? Maybe there are multiple homotopy thinkable, but then what are one or two of them?

It's like saying $f(1)=1$ and $f(2)=4$. Done. Well, to a newbie, maybe secondary school kids, it might be nice to give them an example be it linear $f(x)=ax+b$ or $f(x)=x^2$. To the eyes of the experienced, it might make perfect sense, a specific homotopy might pop out in their minds like popcorns but not in mine.

It's an utter nightmare. I know this is "abstract" math but can not some more specific-ness be put into it? Pictures and diagrams perhaps?

This is only the tip of the massive massive confusion and dumb-foundedness I am experiencing in this area of study. Maybe once it "clicks" it all goes down like an avalanche but so far it's nothing but counter-intuitive.

Can someone please make this possible for me to digest? Suggestions to good websites with examples and diagrams and extensive explanations are also welcome. Thank you

Imagine a malleable object like a rubber band that can occupy different shapes/configurations in space.

Let us call the abstract reference rubber band $X$ (independent of any specific configuration it might take on), and call the space it moves in $Y$.

To define a particular configuration of the rubber band, one would need to specify where each point in the reference rubber band exists in space. That is, each possible configuration of the rubber band in space corresponds to a map from $X$ to $Y$.

Maps $f,g:X \rightarrow Y$ then correspond to two different configurations of the rubber band in space.

Maps $h:X \times I \rightarrow Y$ correspond to a time-sequence of configurations that the rubber band takes on. For every point $x$ in the reference rubber band and for every time $t$ in $I$, the location of that part of the rubber band in space at that time is specified by $h(x,t)$. Visualize $h$ as a movie of the rubber band moving and deforming in space over time.

Now, nothing written so far prevents crazy stuff from happening like the map $f$ taking the rubber band and mapping it to some weird disconnected dust or something like that. Similarly, nothing yet written prevents the rubber band in the movie $h$ to spontaneously teleport from one place to another at every frame, or things like that. The continuity assumptions are minimal assumptions there to prevent such crazy things.

Note that in particular, the continuity assumption still allows the rubber band to pass through itself, or to squish down from a 3d volumetric band to a 2D ring, so long as these processes are done in a continuous manner. If one desires to prevent these cases as well, then more stringent requirements can be used (e.g., isotopy, where the framework is the same but the maps are also required to be injective)

A natural question is, given two continuous configurations $f$ and $g$ of the rubber band, is it possible to continuously move/bend/twist/squish one into the other via a continuous movie $h$? This is equivalent to asking whether there exists a homotopy $h$ between configurations $f$ and $g$.

Whether or not two configurations $f$ and $g$ of an object are homotopic depends on both the shape of the object, and the shape of the ambient space it moves in. Consider a loop on the surface of a donut - in some configurations the loop can be continuously collapsed to a point, whereas in other configurations it cannot. Hence the target of study when using homotopy is often to understand the ambient space that the object moves in, rather than the object itself.

Hope this helps with the intuition.

I'm surprised noone presented a drawing. A homotopy (between paths) can be naively represented as follows.

Suppose we have a path $c_0$ in space. Like this:

We now move it around, indexing its movement on the subscript. For instance,

Making this a continuous process, we get a function $G:I \rightarrow \mathcal{F}(I, Y); \quad s \mapsto c_s$, which induces in a clear manner a function $F: I \times I \rightarrow Y$ by $(s,t) \mapsto c_s(t)$.

A homotopy between maps is simply not restricting ourselves to paths. We consider the maps we want to deform to be maps $f: X \rightarrow Y$, and we get a function $G: I \rightarrow \mathcal{F}(X, Y)$ which induces $F:X \times I \rightarrow Y$ in the same way.

I sympathize with your situation - I also struggled a lot with algebraic topology. I'll try to help you answering specific questions.

So, I am given two sets $X,Y$ whatever they are and two maps $f,g$ that takes an element of $X$ to an element of $Y$. I don't know if $f,g$ are bijective, only injective or surjective or whatever. No information on that. Just maps.

Since we're in algebraic topology, you can safely assume that $X$ and $Y$ are topological spaces, and $f$ and $g$ are continuous maps. We really don't have information about injectiveness/surjectiveness of the maps involved. We can define "homotopy" in another contexts, such as linear algebra, group theory, etc, too, but I'll get to that later.

And this "homotopy" is a ... "map between maps"? And even if so, what $exactly$ is it doing? All the explanation seems to be done with this single line $$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ but no, I don't get what's happening. So $X$ has a bunch of elements $x$, and the product space $X \times I$ gives me elements of the form $\{x,t\} \in X \times I$. Okay. But this map $h$ qualifies as a homotopy as long as the above holds? Then why not just always define $h$ as $h(x,t)=f(x)$ for any $t \neq 1$ and $h(x,t)=g(x)$ for $t=1$, just as how we might define a piecewise function? Then I can define this "homotopy" on $any$ maps.

If you define $h$ as you said above, most likely it will not be continuous at the pairs $(x,1)$, we have a "jump" in $t$ near that. I'd guess that the lecturer there probably said something along the lines of "in this course, unless otherwise stated, all maps will be continuous". So it is not true that there is a homotopy between any maps.

I will give three examples of homotopies and then try to give the general idea:

• Let $f,g\colon X \to \Bbb S^2$ be continuous maps such that $f(x) \neq -g(x)$ for all $x \in \Bbb S^2$, where $X$ is any topological space. Define $h\colon X \times I \to \Bbb S^2$ by $$h(x,t) = \frac{(1-t)f(x)+tg(x)}{\|(1-t)f(x)+tg(x)\|}.$$We are not dividing by zero because of the $f(x) \neq -g(x)$ part (the problem would be for $t = 1/2$). Check that this is a homotopy to get the feeling for the computations (i.e., $h(x,0) = f(x)$, $h(x,1) = g(x)$, and $h$ is continuous).

• Let $f\colon \Bbb R^n \to \Bbb R^n$ be a continuous map. Then $h\colon \Bbb R^n \times I \to \Bbb R^n$ given by $h(x,t) = tx + (1-t)f(x)$ is a homotopy between $f$ and the identity function (check that, same as above).

• Let $f\colon \Bbb R^n \to \Bbb R^k$ be a continuous maps. Then $h\colon \Bbb R^n \times I \to \Bbb R^k$ given by $h(x,t) = tf(x)$ is a homotopy between $f$ and the zero map.

A homotopy is a map between maps in the following sense: fix $f,g\colon X \to Y$ continous maps, and $h\colon X \times I \to Y$ the homotopy. We can see $h$ as the map that takes $t \in I$ to $h_t \in {\cal C}(X,Y)$, where $h_t$ is defined by $h_t(x) = h(x,t)$. Going up a bit ahead, we'll say that a path in a topological space is a map defined on $I$ taking values in the space. Then the above interpretation sees $h$ as a path in ${\cal C}(X,Y)$, connecting $f$ and $g$, since $h_0 = f$ and $h_1 = g$, as functions.

A "dual" interpretation is seen by fixing $x$, and seeing $h$ as the map that takes $x \in X$ to the map $h_x \in {\cal C}(I,Y)$, given by $h_x(t) = h(x,t)$. Then for every fixed $x$, $h_x$ is a path in $Y$, connecting $f(x)$ and $g(x)$.

Here are some notes from where I took the course (in portuguese, but I want a specific image there) - in the first page, there is a square with $f$ in the bottom, $g$ in the top, the left side labeled $I$, etc. The maps $h_t$ are constant in horizontal lines, and the $h_x$ are constant in vertical lines, try to think a bit about that.

Sure, it adds "deforms $f$ CONTINUOUSLY to $g$" but how is that stated in the definition itself? I see it nowhere.

Saying that the deformation is continuous is just jargon for saying that $h\colon X \times I \to Y$ is continuous (we consider the product topology in $X \times I$ here). If the lecturer made the observation of all maps being continuous otherwise stated, as I speculated above, then I think this issue here is solved.

Well, first off what is a "path"? Intuition also doesn't make sense because when mapping one element to another, how can there be different "paths"? $x$ goes to $y$. Done. It's not like going from England to Singapore via either Amsterdam or Frankfurt (thus different paths) is it? Unless it's a map $X \rightarrow Z \rightarrow Y$ and telling me that $x \in X$ goes to $z_1 \in Z$ then to $y \in Y$ or $z_2 \in Z$ and then to $y \in Y$, that might be different paths from $x$ to $y$. But here, it is talking about $X$ and $Y$ only.

As I rushed up before: a path in a topological space $X$, connecting $x$ and $y$ in $X$, is a continuous maps $\gamma\colon I \to X$ such that $\gamma(0) = x$ and $\gamma(1)=y$. It is possible that given $x,y \in X$, there is more than one path connecting them, or that there is no path connecting them, at all. Given continuous maps $f,g$ between topological spaces, there could be more than one homotopy between them, or no homotopy at all.

In general topology, you should've met the concept of a "connected space". We'll say that a topological space is "path-connected" if any two points can be connected by (at least) one path. Every path-connected space is connected, the converse is false. The idea of algebraic topology, at a first moment, is to distinguish topological spaces (modulo homeomorphism) by topological invariants - and the focus is on connectedness/path-connectedness, and one of the most useful instruments to do that depends on homotopies.

Concrete examples of paths:

• given two points in $\Bbb R^2$, we can join them by a straight line, or by a half-circumference, etc.

• if $X = [0,1] \cup [2,3]$, and if $x \in [0,1]$, $y \in [2,3]$, there is no continuous path between $x$ and $y$ - by the intermediate value theorem, such a path $\gamma$ would have to assume the value, say, $1.5$, which is not in $X$ (the problem is that the path passes outside $X$, and we do not want that).

• In $\Bbb S^1 = \{e^{it} \mid t \in \Bbb R\}$, if we take $x = e^{it_0}$ and $y = e^{it_1}$, then $\gamma\colon I \to \Bbb S^1$ given by $\gamma(t) = e^{i((1-t)t_0+tt_1)}$ verifies $\gamma(0) = x$ and $\gamma(1) = y$. This is one possible path between them, going in one direction in the circle - you could go the other way.

Take $X={x}$ the space with a single element $x$. Then a map $f:X→Y$ is the same as and element $f(x)∈Y$. A homotopy $h:f≃g:X→Y$ is the same as a path $h:I→Y$ with initial point $h(0)=f(x)$ and terminal point $h(1)=g(x)∈Y$. A homotopy $h:f≃f:X→Y$ is the same as a closed path $h:I→Y$.

This is the interpretation with $h_x$ that I gave above, but since there only is one $x \in X$, they abused notation and wrote $h\colon I \to Y$, where more precisely we'd write $h_x\colon I \to Y$.

I hope this helps a bit, even. And you can ask away stuff in the comments.

Extra: homotopies in another contexts;

• In linear algebra: given vector spaces $V,W$, and linear maps $T,S\colon V \to W$, a homotopy between $T$ and $S$ is a map $H\colon V \times I \to W$ such that $H(x,0) = T(x)$, $H(x,1) = S(x)$, and for all $t \in I$, the maps $H_t(x) = H(x,t)$ is linear;

• In group theory: given groups $G,H$, and group homomorphisms $\varphi, \psi\colon G \to H$, a homotopy between $\varphi$ and $\psi$ is a map $h\colon G \times I \to H$ such that $h(x,0) = \varphi(x)$, $h(x,1) = \psi(x)$, and for all $t \in I$, $h_t(x) = h(x,t)$ is a group homomorphism;

And so on. If you know a bit of category theory, you see the pattern: we want a $1$-parameter family of morphisms joining the initial ones. For topological spaces, a $1$-parameter family of continuous maps; for vector spaces, a $1$-parameter family of linear maps, etc.

There's a pedagogical quirk that makes homotopies extremely easy to explain, and that's that they can (should?) be thought of as occurring over time, and there's this wonderful invention called a video that very naturally expresses how shapes can change over time.

Try Regular homotopies in the plane... it's some sort of cult classic but I recommend it with no irony for your question.