First suppose that \([f] = 0\) in \(\pi_n(X, A, x_0)\text{.}\) This means that we have a homotopy of triples
\begin{equation*}
H \colon (D^n \times I,\; S^{n-1} \times I,\; \{s_0\} \times I) \to (X, A, x_0)
\end{equation*}
from \(f\) to the constant map at \(x_0\text{.}\) Using \(H\) we define another homotopy \(H'\) relative to \(S^{n-1}\) is defined by
\begin{equation*}
H'(x,t) =
\begin{cases}
H\Bigl(\dfrac{x}{1 - t/2},\; t\Bigr), & 0 \leq \lvert x \rvert \leq 1 - \dfrac{t}{2}, \\[10pt]
H\Bigl(\dfrac{x}{\lvert x \rvert},\; 2 - 2\lvert x \rvert\Bigr), & 1 - \dfrac{t}{2} \leq \lvert x \rvert \leq 1.
\end{cases}
\end{equation*}
Conversely, suppose \(f\) is homotopic relative to \(S^{n-1}\) to a map \(g \colon D^n \to X\) with \(g(D^n) \subseteq A\text{.}\) Since the homotopy fixes \(S^{n-1}\text{,}\) it is a homotopy of triples, so \([f] = [g]\) in \(\pi_n(X, A, x_0)\text{,}\) and \(g(s_0) = f(s_0) = x_0\text{.}\) Now contract \(D^n\) to \(s_0\) along straight lines:
\begin{equation*}
K(x, t) = g\bigl((1-t)x + t s_0\bigr).
\end{equation*}
This is a homotopy of triples from \(g\) to the constant map \(c_{x_0}\text{:}\) all its values lie in \(g(D^n) \subseteq A\text{,}\) so certainly \(K(S^{n-1} \times I) \subseteq A\text{,}\) and \(K(s_0, t) = g(s_0) = x_0\) for all \(t\text{.}\) Hence \([f] = [g] = [c_{x_0}] = 0\text{.}\)
