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?


1 Answer 1


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.

  • $\begingroup$ I think you should add some details to provide an answer. You should point out that $H_\omega$ neither starts on $f_0\omega\ast H_y$ nor finishes on $H_x\ast f_1\omega$, so must be further homotoped. Neither is $H_\omega$ a path homotopy (since $H_\omega(0,t)\neq f_0(x)$, $H_\omega(1,t)\neq f_1(y)$ in general). I was under the impression that the OP was hoping for some intuition for the required homotopies to produce from $H_\omega$ the correct path homotopy. $\endgroup$
    – Tyrone
    Mar 4, 2018 at 10:31
  • $\begingroup$ Yes, @Tyrone I agree with you. Do you know how to define an adequate homotopy? Or maybe some sort of argument which leads to the conclusion that $f_0\omega \ast H_y$ is path homotopic to $H_x\ast f_1\omega$ by considering general properties. $\endgroup$ Mar 4, 2018 at 11:44
  • $\begingroup$ @tyrone Thanks for the suggestion. I misunderstood where the difficulty was. $\endgroup$ Mar 4, 2018 at 17:27
  • $\begingroup$ Sorry... I think I did not understand. If $T=H_{\omega}$ then $a=H(\omega(0),t)=H_x(t)$ and similarly $b=H_y(t)$, $c=f_0\omega$, $d=f_1\omega$. So $d\ast a=f_1\omega \ast H_x$ and $b\ast c=H_y\ast f_0\omega$, and that is not what I want. Moreover, is there an explicit formula for the $f$ you defined? It is very complicated to me to follow the "visual" definition, maybe if I had the formula I would be able to prove the assertion in an analytical way, which best suits to me. $\endgroup$ Mar 4, 2018 at 20:43
  • $\begingroup$ Sorry about that. It's easy to fix, if I misstated something, but something must be wrong-$f_1\omega *H_x$ and $H_x *f_1\omega$ can't both be defined. Think back through your definitions-it seems to me that the latter is the issue, unless we're just notating composition of paths in different orders. There certainly is an explicit formula for the function I defined, which you should write down for yourself. I've described the function as a composition of a projection, rotation, translation, and scaling, all of which admit explicit formulas I presume are well known to you $\endgroup$ Mar 4, 2018 at 22:40

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