Constructing homotopy

Question

Let $X,Y$ be topological spaces and $f,g:X \rightarrow Y$ be continuous maps. Suppose $f,g$ are homotopic by a homotopy $H:X \times [0,1] \rightarrow Y$. Let $\gamma$ be a path in $Y$ with endpoints $y_0=f(x_0)$ and $z_0=g(x_0)$ defined by $\gamma(t)=H(x_0,t)$.
Let $\alpha:[0,1] \rightarrow X$ be a loop in $X$ with $\alpha(0)=\alpha(1)=x_0$.

I would like to show that the paths $f \circ \alpha$ and $\gamma \ast ((g \circ \alpha) \ast \bar{\gamma})$ are path homotopic by constructing a homotopy $G:[0,1] \times [0,1] \rightarrow Y$.

I know that $\gamma$ induces a change of base point isomorphism $T_{\gamma}:\pi_1(Y;z_0) \rightarrow \pi_1(Y;y_0)$ given by $T_{\gamma}([\beta])=[\gamma \ast \beta \ast \bar{\gamma}]$ for all $[\beta]\in \pi_1(Y;z_0)$.

I don't know how to build this homotopy $G$ specifically, but I know it must satisfy the following:
$$G(s,0)=(f \circ \alpha)(s)$$
$$G(s,1)=(\gamma \ast ((g \circ \alpha) \ast \bar{\gamma}))(s)$$
$$G(0,t)=(f \circ \alpha)(0)=(\gamma \ast ((g \circ \alpha) \ast \bar{\gamma}))(0)$$
$$G(1,t)=(f \circ \alpha)(1)=(\gamma \ast ((g \circ \alpha) \ast \bar{\gamma}))(1).$$

Answer 1

THEOREM: Let $h:X \to Y$ be a continuous function. If $α,β$ are homotopic paths in $X$ then $hoα$ and $hoβ$ are homotopic in $Y$.
We're given that there exists a continuous map $H:X×I \to Y$ such that $H(x,0)=f(x)$ and $H(x,1)=g(x)$.
Hence in order to show that paths $foα$ and $\gamma \ast (goα) \ast \hat \gamma$ are homotopic we have to look for paths in $X×I$ such that their image under H is $foα$ and $\gamma \ast (goα) \ast \hat \gamma$.
Let $c(t)=(x_0,t)$, $p(s)=(α(s),0)$, $q(s)=(α(s),1)$. Hence $p$ and $c \ast q \ast \hat c$ are our required paths.
Now let $G:I×I \to X×I$ be defined as $G(s,t)=(α(s),t)$. Now if we can find paths in $I×I$ such that their image under G are $p$ and $c \ast q \ast \hat c$ we're done because that means $p$ and $c \ast q \ast \hat c$ are homotopic.
Let $a(t)= (0,t)$, $b(s)=(s,1)$, $c(t)=(1,t)$ and $d(s)=(s,0)$.
It can be seen that $d$ and $a \ast b \ast \hat c$ are the required paths. Now since both these paths $d$ and $a \ast b \ast \hat c$ have the same initial and terminal point and since $I×I$ is simply connected they are homotopic. Implies $p$ and $c \ast q \ast \hat c$ are homotopic implies $foα$ and $\gamma \ast (goα) \ast \hat \gamma$ are homotopic.