Let $X$ be a top space and $A \subset X$ a path-connected deformation retract. I.e., there exists a homotopy
$$H:X \times [0,1] \to X$$ such that for every $x \in X$ one has $H(x,0)=x$ and $H(x,1) \in A$. And $H(a,t)=a$ for every $a \in A$. For sake of avoiding trivialities, let us suppose $y,z \in X \setminus A$. We wish to construct a path between the two points. Note that
$$H(y,0)=y, H(z,0)=z$$ and $$H(y,1), H(z,1) \in A$$
denote them $a_1,a_2$ then as $A$ is path-connected, there exists some $$\gamma:[0,1] \to A$$
with $\gamma(0)=a_1, \gamma(1)=a_2$. So $H$ is a map from $y \in X$ to $a_1 \in A$ and also a map from $z \in X$ to $a_2 \in A$ and $\gamma$ sends $a_1$ to $a_2$ So to I take the homotopy of $H$ circ with $\gamma$ circ with $H^{-1}$? So im trying to take two points in $X$ and send one of them to $A$ then link it with the image of the other under the homotopy then map it back to $X$ to stop at $z$.
Take $y,z\in X$ and let $a_y=H(y,1)$ and $a_z=H(z,1)$. Since $A$ is path connected, there is a path $\gamma\colon[0,1]\longrightarrow A$ such that $\gamma(0)=a_y$ and the $\gamma(1)=a_z$. Now, let $\gamma_y(t)=H(y,t)$ and let $\gamma_z(t)=H(z,t)$ ($t\in[0,1]$). Then $\gamma_y$ and $\gamma_z$ are paths, $\gamma_y(0)=y$, $\gamma_y(1)=a_y$, $\gamma_z(0)=z$, and $\gamma_z(1)=a_z$. So, $\gamma_y*\gamma*\gamma_z^-$ (where $*$ denotes concatenation) is a path in $X$ going from $y$ to $z$.