EDIT: I was reading little bit of homotopy theory in trying to understand the difference between homotopic maps from $X\to Y$ and homotopic paths in $Y$, and their significance in the context of SU(2) Yang-Mills instantons.

A. For SU(2) Yang-Mills instantons, we have maps from $X\to Y$ i.e., $\mathbb{S}^3\to \mathbb{S}^3$. Here, $X=\mathbb{S}^3$ corresponds to the boundary of the four-dimensional Euclidean space, and $Y=\mathbb{S}^3$ corresponds to the SU(2) group manifold. Two different maps $f^{(m)}(x)$ and $f^{(n)}(x)$ from $\mathbb{S}^3\to \mathbb{S}^3$, with winding number $m$ and $n$ respectively, are not homotopic for $m\neq n$. Technically, this means, one cannot find a homotopy $H(x,t)$ from $X\times [0,1]\to Y$ which can continuously deform the map $f^{(n)}(x)$ to $f^{(m)}(x)$ in the sense $$H(x,0)=f^{(m)}(x)\hspace{0.2cm} \text{and}\hspace{0.2cm} H(x,1)=f^{(n)}(x).$$ Here, we are talking about maps from $X\to Y$. It turns out that the maps from $\mathbb{S}^3\to \mathbb{S}^3$ can be classified as $$\pi_3[\mathbb{S}^3]=\mathbb{Z}.$$

B. As I understand paths in $Y$, are not same as $f(x), g(x)$ defined above. Because a path $a(s)$ is defined in terms of a parameter $s\in[0,1]$ and is defined without any reference to what X is. It is possible to see whether the space Y has more than one class of paths without reference to $X$, and paths are not maps from $X\to Y$ but from $[0,1]\to Y$.

It said that non-homotopic maps from $\mathbb{S}^n\to Y$ are possible only if $\pi_n[Y]$ (where $Y$ is any manifold) is non-trivial (and in our case it is indeed so). Is it a theorem? Apparently I don't see any connection between A and . It is not clear to me why non-homotopic maps from $X\to Y$ will not be possible if $\pi_n[Y]$ is trivial.

I don't have much understanding of topology or homotopy theory, and I'm not sure whether the question makes sense. If not I'll try to clarify it further. If possible, a not-too-technical answer will be helpful for me.

  • $\begingroup$ Why do you say that the number of inequivalent class of paths in $Y$ gives the number of non-homotopic maps $X\to Y$? Also, notice that $\pi_n(Y)$ is, by definition, the space of maps $S^n\to Y$ up to homotopy. $\endgroup$
    – coconut
    Jan 26, 2017 at 12:01
  • $\begingroup$ I have corrected it. While defining any path in Y, it is immaterial what the X is. Right? But to classify paths into inequivalent homotopy classes, one need to know what X is. Is that right? But this is not clear to me. I understand $\pi_1[Y]$ as set of all closed paths in Y, which can be continuously deformed to a point. How is this related to a map from $S^1\to Y$? I have the impression that maps are different than paths. $\endgroup$
    – SRS
    Jan 26, 2017 at 12:22
  • 1
    $\begingroup$ The closed paths in the $\pi_1(Y)$ don't need to be contractible! About the second question: a closed path is a map from an interval to $Y$ for which the image of the endpoints coincide. Because of this, you might just identify the endpoints of interval, getting the circle $S^1$ $\endgroup$
    – coconut
    Jan 26, 2017 at 12:32
  • 1
    $\begingroup$ The third homotopy group does not classify classes of loops. It classifies classes of spheres $S^3$. As I understand it checkes wheter different hypersurfaces can be deformed into each other. $\endgroup$
    – Diracology
    Jan 26, 2017 at 12:34
  • 1
    $\begingroup$ @SRS If you care about the space of maps from the three-sphere to your space and their homotopy properties, then you care about the third homotopy group, because they are the same thing! Of course, if you are dealing just with a couple of maps, you don't need the full space, but the information that homotopy theory gives you might be useful (for example, the case in which it is known that the homotopy group is trivial, you don't need more calculations to know that all the maps from the nth sphere are trivial) $\endgroup$
    – coconut
    Jan 26, 2017 at 13:29

1 Answer 1


A definition might clarify the confusion. The $n$-th homotopy group $\pi_n(Y)$ of a space $Y$ is constructed using the set of maps $S^n\to Y$. If we identify two maps when they're homotopic and add some group structure (which we don't need for this answer) we get $\pi_n(Y)$. It is then clear that, whenever $\pi_n(Y)$ is trivial (has one element) any two maps $S^n\to Y$ will be homotopic.

The connection with the space of paths can only be made for the case $n=1$, in which $\pi_1(Y)$ (the set of maps from the circle to $Y$ up to homotopy, which is known as the fundamental group) is the space of closed paths in $Y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.