I am currently studying undergraduate algebraic topology. We have in our notes the definition of homotopy of continuous maps:

Two continuous maps $\phi_0,\phi_1:X \rightarrow Y$ are said to be homotopic if there exists a continuous map $F:X \times I \rightarrow Y$ (where $X \times I$ has the product topology) such that

$$F(x,0) = \phi_0(x), \; F(x,1) = \phi_1(x), \; \forall x \in X$$

This definition makes sense to me. I am now trying to gain a more intuitive understanding.

For comparison, we have previously had the definition of based homotopy of paths in a topological space $X$. This is relatively easy to understand - two paths with the same end points are homotopic if you can deform one into the other in a continuous manner.

I'm getting mixed up, though, over how a continuous map, say $\phi_0$, from a space $X$ into a space $Y$ can be homotopic in such a sense to another such map, $\phi_1$. If things have a Euclidean topology, what sort of continuous maps are not homotopic?

If for example, $X = Y = T^2, \; \phi_0 = id_x$, what might be some examples of continuous maps $\phi$ which $\phi_0$ is or is not homotopic to?

Is there a nice way to visualise this, in a similar manner to how we can visualise based homotopy of paths?

  • $\begingroup$ In your example, $\phi_0$ is not homotopic to a constant map, since the torus is not contractible. Actually though, a lot of homotopy theory is basically answering this question $\endgroup$ – leibnewtz Nov 13 '17 at 22:23
  • $\begingroup$ Your mention of "if things have a Euclidean topology" suggests to me that maybe you're not thinking of the general situation of maps $X\to Y$ but of the special case where $Y$ is a Euclidean space (not a subspace like a circle but the whole $\mathbb R^n$). In that special case, it is true that any two continuous maps $\phi_0,\phi_1:X\to\mathbb R^n$ are homotopic. $\endgroup$ – Andreas Blass Nov 13 '17 at 23:41

This is in fact quite a hard question in general (determining whether two maps are homotopic), and as leibnewtz says much of algebraic topology is devoted to answering it in one form or another. Here is perhaps the simplest example: consider the unit circle $S^1$, thought of as the set of all complex numbers of norm $1$. I claim that none of the maps

$$S^1 \ni z \mapsto z^n \in S^1, n \in \mathbb{Z}$$

are homotopic to each other. This is perhaps the simplest interesting example in algebraic topology and can be understood from a variety of different perspectives; for more information see winding number. For visualization purposes it may help to replace the second copy of $S^1$ with the punctured complex plane $\mathbb{C} \setminus \{ 0 \}$, so we're talking about deforming loops in $\mathbb{C} \setminus \{ 0 \}$ into other loops.


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