A natural transformation between the categories $\pi_1X$ and $\pi_1Y$ Let $\mathscr{C}$ and $\mathscr{D}$ be categories and let $F_0,F_1$ be covariant functors $\mathscr{C}\to \mathscr{D}$. A natural transformation $\alpha:F_0\to F_1$ is a collection $\alpha=\left \{\alpha_A:A\in \text{obj } \left (\mathscr{C}\right )\right \}$ such that $\alpha_A\in \hom_{\mathscr{D}}\left (F_0A,F_1A\right )$ and for every $f\in \hom_{\mathscr{C}}\left (A,B\right )$ we have $F_1\left (f\right )\alpha_A=\alpha_BF_0\left (f\right )$.
If $X$ is a topological space, then $\pi_1X$ is a grupoid. It is a category whose objects are the elements of $X$ and the morphisms between two objects $x,y\in X$ are the paths that join $x$ to $y$ modulo path homotopies. Therefore, every morphism is an equivalence.
If $f:X\to Y$ is a continuous function, then $f_{\ast}:\pi_1X\to \pi_1Y$ is a functor which sends every $x\in X$ to $f\left (x\right )$ and every $\left [\alpha\right ] \in \hom_{\pi_1X}\left (x,y\right )=:\pi_1\left (X;x,y\right )$ to $\left [f\alpha\right ]\in \pi_1\left (Y;f\left (x\right ),f\left (y\right )\right )$. Observe that if $x=y$ then we have the usual group homomorphism $f_{\ast}:\pi_1\left (X,x\right )\to \pi_1\left (Y,f\left (x\right )\right )$.
I have to prove the following assertion:

Let $f_0,f_1:X\to Y$ be continuous functions and let $H:X\times I\to Y$ (where $I=\left [0,1\right ]$) be an homotopy between $f_0$ and $f_1$. In other words, $H$ is continuous and $H\left (x,i\right )=f_i\left (x\right )$ for every $i\in \left \{0,1\right \}$. Then $H$ induces a natural transformation $\left (f_0\right )_{\ast}\to \left (f_1\right )_{\ast}$.

This is my attempt, it is not concluding: For every $x\in X$ define $H_x:I\to Y$ such that $H_x\left (t\right )=H\left (x,t\right )$. Then $H_x\left (i\right )=f_i\left (x\right )$ for every $i\in \left \{0,1\right \}$. Therefore $\left [H_x\right ]\in \pi_1\left (X;f_0\left (x\right ),f_1\left (x\right )\right )$.
I want to prove that $\left \{\left [H_x\right ]:x\in X\right \}$ is our natural transformation. In order to prove that, we take $x,y\in X$ and $\left [\omega\right ]\in \pi_1\left (X;x,y\right )$.
I would be done if I were able to prove that $H_x\ast f_1\omega$ is path homotopic to $f_0\omega \ast H_y$, but at this step I got stuck. How would you prove that those functions are path homotopic?
 A: There is already a canonical such homotopy given, namely $H_\omega:I\times I\to Y$, where to be clear $H_\omega(s,t)=H(\omega(s),t)$. This is a square whose boundary consists of the four paths at issue. So it suffices to prove the following: from any map $T:I\times I\to X$ with $T(0,t)=a,T(1,t)=b,T(s,0)=c,T(s,1)=d$ induces a path homotopy between $d*a$ and $b*c$. This can be done via the map $f: I\times I  \to I\times I$ which views $I\times I$ as the unit square in the first quadrant of the plane, projects vertically onto the square with vertices $(1/2,0),(1/2,1),(0,1/2),(1,1/2)$, then maps the later square back onto $I\times I$ by rotating counterclockwise through an angle of $\pi/4$ then translating and scaling.
Then $T\circ f$ is constant on its left and right edges, and runs through $d*a$ and $b*c$ on the top and bottom, giving the desired homotopy endpoints fixed.
The point of this unpleasantly explicit construction is to show how to transform any map from a square into a homotopy between the composed paths on its boundary. A more abstract way to say this is that there is a self-homotopy equivalence of the square which induces a homeomorphism of the square with $I\times I/(0\times I\sqcup 1\times I)$, the latter being the space which represents path homotopies, acting in the desired way on boundaries. 
