TITLE: Explain "homotopy" to me
QUESTION [96 upvotes]: 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

REPLY [6 votes]: So the first thing that is going on is that somewhere in your lecture notes or in your text book, they defined "map" to mean "continuous function".
Now, I only know this because your quotes won't make sense if that isn't true.  So I work backwards from the definitions you are providing me to work out the definitions of the terms used in your definitions.  This is sub-optimal, but mathematics is often full of jargon: being able to guess at jargon helps.
A good way to approach this problem is whenever someone uses a term that isn't the totally typical one (ie uses map instead of function), put the words "sufficiently nice" in front of it, and suspend judgement on how nice it needs to be until later.  You'll sometimes guess wrong, but it helps offload the work required to understand the definition until later.
And "nice" basically means "has just barely the properties needed to make the definition/theorem true, in a way that is consistent with the other theorems and definitions nearby".
This relies on the fact that mathematicians are lazy, and define away problems when they run into them.

Now, let's apply this guess.

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.

and it adds

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

Replace map with continuous function:

A homotopy between continuous functions $f,g:X \rightarrow Y$ is a continuous function $h: X \times I \rightarrow Y$ such that
$$h(x,0)=f(x), h(x,1)=g(x) \in Y$$ where $x \in X$ and $I=[0,1]$. We say continuous functions $f,g$ are homotopic.

and it adds

A homotopy deforms the continuous function $f$ continuously to $g$.

which makes things make a bit more sense.
Next, I'll deconstruct this:

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

So $h|_{X \times \{0\}} = f$ and $h|_{X \times \{1\}} = g$.  $h(?,0)$ is $f(?)$ and $h(?,1)$ is $g(?)$.
(For some function $z$, $z|_{Set}$ is $z$ restricted to $Set$, ie if we only talk about the part of $z$ that is on $Set$, what function do we get?  Here I'm lazy again, and despite the fact that $h|_{X \times \{0\}}$ is a function from $X \times \{0\} \rightarrow Y$ and $f$ is a function from $X \rightarrow Y$, I say they are equal, because $X \times \{0\}$ is basically $X$.)
And, most importantly, it is continuous -- so its behavior at the end points restricts (to some degree) what it does in the middle!
The existend of a homotopy means $f$ and $g$ are homotopic.  Often we do not care what the homotopy looks like, merely that it exists.
If it exists, then we can "continuously deform" $f$ into $g$ by looking at $h(?,t)$ where $t$ goes from $0$ to $1$.  When $t=0$ we get $f$, when $t=1$ we get $g$, and in the middle we get a continuous deformation between the two.
You can think of $h:I \rightarrow (X \rightarrow Y)$, but that requires a friendly definition of continuity on continuous functions from $X \rightarrow Y$, and it might lead to some complications.  $h$ in a sense maps $I$ (the unit interval) to a continuous deformation of functions that starts at $f$ and ends at $g$.


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 $h: I \rightarrow Y$.

Now here we see a mathematican being lazy.  If you have the space $I \times \{x\}$, that is basically the same space as $I$, because the cross product of a space with a set containing a single element is basically the same as the space.  What the single element is is mostly unimportant.
So they silently and without comment drop the $\{x\}$.  Lazy mathematician.
A "path" from a point $a$ to a point $b$ both in $Y$ is just a map $p : I \rightarrow Y$ where $p(0) = a$ and $p(1) = b$.  And remember a map is a continuous function.
Each path from England to Sinapore is a different path.  Go north 2 km, the strait to England?  Yep.  Travel in a spiral around the world 5 million times, then stop in England?  Yep.  Go to the moon?  Yep.  Travel into a black hole, exit out a white hole and then swing around to England?  Yep.  Use the cantor set so you only move on a set of measure zero and end up in England?  Yep.  All (cateogies of) paths to England.
But do remember that we don't have differentiability: just continuity at this point.  Paths can be non-differentiable.
A path from 1 to 2 in R might be $f:I \rightarrow R$ such that $f(x) = x+1$.  Another is $f(x) = x^2+1$.  Or you could use the cantor set to define the function.
In topology, it is the exitence of any path you more often care about, not the definition of a given path.  We get theorems about paths existing without having to construct them.
Now, what is worse is that pictures cannot work: they can give trivial examples (like the ones above), but a non-differentiable path cannot be drawn: every drawn line is differentiable.