Exercise II.4.1 in Massey's A basic course in algebraic topology

Question

Notation:

• $\Pi_1:\mathbf{Top}\to \mathbf{Grpd}$ the functor which maps a topological space $X$ to the fundamental groupoid $\Pi_1(X)$.
• $\pi_1:\mathbf{Top}_{\bullet}\to\mathbf{Grp}$ the functor which maps a pointed topological spaces $(X,x)$ to the fundamental group $\pi_1(X,x)$.
• Let $X$, $Y$ be topological spaces and $f:X\to Y$ be continuous. Regard $\Pi_1(X)$ as a category with $\operatorname{Obj}(\Pi_1(X))=X$ and $\operatorname{Hom}(x_1,x_2)$ the collection of equivalence classes of pathes in $X$ starts from $x_1$ to $x_2$, so $\Pi_1(f):\Pi_1(X)\to\Pi_1(Y)$ becomes a functor with $\Pi_1(f)(x)=f(x)$ for $x\in X$ and $\Pi_1(f)(\alpha)=f_*(\alpha)=[f\circ g]$ for $g\in\alpha$, $\alpha\in\operatorname{Hom}(x_1,x_2)$.

The morphism-part of the functor $\Pi_1(f)$ is $\require{AMScd}$ \begin{CD} x_1 @.\mapsto@.f(x_1)\\ @V\alpha VV@. @VV f_\ast(\alpha) V\\ x_2 @.\mapsto@.f(x_2) \end{CD}

We have the diagram $\require{AMScd}$ \begin{CD} \pi_1(X,x_1) @> f_\ast >> \pi_1(Y,f(x_1))\\ @V \beta_\alpha VV @VV \beta_{f_\ast(\alpha)} V\\ \pi_1(X,x_2) @>> f_\ast > \pi_1(Y,f(x_2)) \end{CD} Here $\beta_\alpha$ is the change of basepoint homomorphism of $\alpha$.

This two "diagrams" look similar (though the former one isn't a diagram), so I am wondering whether I can prove the later from the former without analyzing the element in $\pi_1(X,x_1)$. For example, if I have a commutative diagram in category $\mathcal{C}$ and a functor $\mathcal{C}\to\mathcal{D}$, then I have a commutative diagram in $D$. But in this case, the former isn't a diagram and I don't have an appropriate functor from $\mathbf{Grpd}$ to $\mathbf{Grp}$.

The original proof of the later diagram is by direct calculation:

Let $g:I=[0,1]\to X$ be a loop based at $x_1$ and $\gamma$ denote the equivalence class of $g$. Let $h$ be a path belonging to the path class $\alpha$.

Then \begin{align*} \beta_{f_*(\alpha)}( f_*(\gamma))&=\overline{f_*(\alpha)}\ast[f\circ g]\ast f_*(\alpha)=[f\circ \overline{h}]\ast[f\circ g]\ast [f\circ h]=[f\circ(\overline{h}\ast g\ast h)]\\&=f_*(\beta_\alpha(\gamma)) \end{align*}

Answer 1

I can only guess what you mean. You want to derive (2) from the property that $\Pi_1(f)$ is a functor? At least, we can abstract this with the following Lemma.

Lemma. Let $F : \mathcal{C} \to \mathcal{D}$ be a functor. Let $\beta : x \to x'$ be an isomorphism in $\mathcal{C}$. Then the diagram of groups $$\begin{array}{ccc} \mathrm{Aut}(x) & \xrightarrow{F} & \mathrm{Aut}(F(x)) \\ c_\beta \downarrow ~~~ && ~~~\downarrow c_{F(\beta)}\\ \mathrm{Aut}(x') & \xrightarrow{F} & \mathrm{Aut}(F(x')) \end{array}$$ commutes. Here, $c_\beta$ denotes conjugation with $\beta$.

The proof is a trivial calculation: For $f \in \mathrm{Aut}(x)$ we have $$c_{F(\beta)}(F(f))=F(\beta) F(f) F(\beta)^{-1} = F(\beta f \beta^{-1})=F(c_{\beta}(f)).$$

Applying this to the functor $\Pi_1(f) : \Pi_1(X) \to \Pi_1(Y)$ associated to a continuous map $f : X \to Y$ gives the desired diagram of fundamental groups.