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

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*}

• Can you (briefly) sketch how you proved it element-wisely? May 25, 2021 at 8:52
• 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 $\beta_{f_*(\alpha)}( f_*(\gamma))=\overline{f_*(\alpha)}\ast[f\circ g]\ast f_*(\alpha)=[f\circ \overline{h}]\ast[f\circ g]\ast [f\circ h]=[f\circ(\overline{h}\ast g\ast h)]=f_*(\beta_\alpha(\gamma))$ May 25, 2021 at 9:16
• Could you rephrase your question? And btw (1) is not a commutative diagrams, it's just a mapping. May 26, 2021 at 22:35
• The sentence "Now if X and Y are topological spaces with X path-connected, then we have the functor ..." still does not make much sense. 0) You don't say between which categories you describe a functor. 1) The diagram is not a functor. 2) But actually you just describe the morphism-part of the functor $\Pi(f)$, which has already been mentioned before. Better just say that. 3) You don't need assumption that X is path-connected. Also, your question still refers to a diagram (1), but (1) is not a diagram. That's why I downvote, the question is not clear at all. I can only guess what you mean. May 27, 2021 at 19:28

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.