On the definition of homotopy.

Question

According to the Formal definition section in Wikipedia's article on homotopy a homotopy is defined simply as follows:

Formally, a homotopy between two continuous functionss $f$ and $g$ from a topological space $X$ to a topological space $Y$ is defined to be a continuous function $H: X \times [0,1] \to Y$ from the product topology of the space $X$ with the unit interval $[0,1]$ to $Y$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x \in X$.

Now what I am surprised about this definition is that it looks too weak. To show what I mean, let me introduce an example, and tell me please if something is wrong in my reasoning.

Let $X := \mathbb{R}^2-\{0\}$ be equipped with the subspace topology of the Euclidean topology on $\mathbb{R}^2$, while

$$\gamma_1:[0,1]\to X:t\mapsto (\cos(2\pi t),\sin(2\pi t))\quad \text{and}\quad \gamma_0:[0,1]\to X:t\mapsto \gamma_1(0) = (1,0)$$

Then by the above definition we can define the following homotopy between these:

$$h:[0,1]\times[0,1]\to X:(s,t)\mapsto \gamma_1(st)$$

After all, $h$ is evidently $C^\infty ([0,1]^2,X)$, hence continuous, and $h(0,t) = \gamma_0(t)$ as well as $h(1,t) = \gamma_1(t)$ for all $t\in[0,1]$. Consequently, the constant loop (i.e. a point) is homotopic to the unit circle around the point $(0,0)\notin X$. This feels extremely weird, because it would imply any loops in the punctured plane are homotopic to each other.

I see, nevertheless, that the Definition section on Wikipedia's article about the fundamental group uses a stricter definition that the above map $\gamma$ does not satisfy, namely because for all $s\in]0,1[$ we have $\gamma(s,1)\neq (1,0)$. Hence I have two questions:

1. Is the above map $\gamma$ really a homotopy in the sense given by the first definition, or did I commit a mistake in my reasoning?
2. Since the first definition, by a similar construction to the above example, renders every two curves with a same starting point homotopic (because we can always contract the first one to the initial point and then expand it to become the second curve), what is the point of this so permissive definition of homotopy? Again, if I am missing something out about the first definition, please tell me.

Thank you in advance.

Answer 1

The definition of homotopy you give is correct. This is a fully general definition, that works for all spaces and continuous mappings between them.

When looking at something like the fundamental group of a space $X$, as you correctly point out, it is not interesting to look at whether or not two mappings $[0, 1] \to X$ are homotopic. In fact, you'll find that the equivalence classes of such maps just give you the path connected components of $X$, and we didn't need paths at all.

Of course, we don't make definitions for the sake of definitions. We want to capture the "hole" in the punctured plane. Therefore the definition of the fundamental group usually starts not with all paths $f, g : [0, 1] \to X$, but only with those where $f(0) = f(1) = g(0) = g(1) = x_0$, some fixed point (a basepoint) of $X$. Then, we make the further requirement that our homotopy $H$ is a homotopy relative to $\{0, 1\}$, which means that for any $t$, we furthermore have $H(0, t) = f(0) = g(0)$ and $H(1, t) = f(1) = g(1)$. This gives you the definition you are used to.

Strictly formally speaking, you probably should not simply say that "$f$ and $g$ are homotopic", if you mean "homotopic relative to $\{0, 1\}$", but it gets a bit repetitive, and it is normally quite clear from context which notion of homotopy is meant.

If you want to use the general definition of homotopy directly, you could define the fundamental group of a (path connected) space $X$ directly by looking at homotopy equivalence classes of maps $S^1 \to X$, using the circle instead of the interval. But then defining the group operation on the fundamental group is a lot harder, and this doesn't really have a particular advantage.