Connection between homotopic maps from $X\to Y$ and homotopic paths in $Y$ in the context of SU(2) Yang-Mills instantons

Question

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.

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 $Y$.