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Let $\text{ppTop}$ denote the category of pointed and path connected topological spaces with morphisms base-preserve continuous maps. The fundamental group gives a functor $FG: \text{ppTop}\to \text{Gp}$ where GP is the category of groups.

Now we consider the category $\text{pTop}$ consisting of path-connected topological spaces and we can naturally define the fundamental groupoids instead of fundamental groups on $\text{pTop}$. If we want to define the fundamental group then we need to choose a base point. Notice that there is a forgetful functor $\text{For}:\text{ppTop}\to \text{pTop}$.

My question is: could we lift the functor $FG: \text{ppTop}\to \text{Gp}$ to a functor $\widetilde{FG}: \text{pTop}\to \text{Gp}$ such that $\widetilde{FG}\circ \text{For}=FG$? If not, how to construct a contradiction?

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  • $\begingroup$ There is a canonical way of associating to a space its fundamental group up to conjugacy, however. $\endgroup$ Nov 28, 2019 at 21:57

2 Answers 2

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For any such lift $\widetilde{FG}:\mathrm{pTop}\to \mathrm{Gp}$ the induced map $pTop(X,Y)\to Gp(\widetilde{FG}(X),\widetilde{FG}(Y))$ must factor through the set of homotopy classes of the maps between $X$ and $Y$, for any spaces $X,Y$.

Indeed, consider the maps $\iota_0,\iota_1:X\rightrightarrows X\times[0,1]$ given by $\iota_a(x)=(x,a)$ for $a=0,1$. The morphisms $\iota_0,\iota_1$ are images of the morphism $\tilde\iota_0:(X,x)\to (X\times[0,1],(x,0))$ and $\tilde\iota_1:(X,x)\to (X\times[0,1],(x,1))$ in the pointed category. Since the functor $FG$ carries $\tilde\iota_0$ and $\tilde\iota_1$ to isomorphisms, so does the lift $\widetilde{FG}$ to the morphisms $\iota_0,\iota_1$. Analogously, the projection $p:X\times[0,1]\to X$ gets sent to an isomorphism. Since $p\circ\iota_0=p\circ\iota_1=id_X$, the induced morphisms $\widetilde{FG}(\iota_0),\widetilde{FG}(\iota_1):\widetilde{FG}(X)\to \widetilde{FG}(X\times[0,1])$ must be equal. In particular, the two compositions $$pTop(X\times[0,1],Y)\rightrightarrows pTop(X,Y)\to Gr(\widetilde{FG}(X),\widetilde{FG}(Y))$$ are equal which implies that any two homotopic maps $f_0,f_1:X\to Y$ induce the same maps between $\widetilde{FG}(X)$ and $\widetilde{FG}(Y)$.

Take now $X=S^1$ equipped with a base point $p\in S^1$. For any $(Y,y)\in ppTop$ the map induced by $FG$ sends a pointed morphism to its homotopy class$$ppTop((S^1,p),(Y,y))\twoheadrightarrow Gp(\mathbb{Z},\pi_1(Y,y))=\pi_1(Y,y)$$

By the above observation, a lifting $\widetilde{FG}$ would yield a factorization of this map through the set of homotopy classes of unpointed maps $S^1\to Y$.

[Corrected on 11/29 thanks to a comment by Achim Krause]:

However, the latter set is identified with the set of conjugacy classes in $\pi_1(Y,y)$, so taking $Y$ to be any space with a non-abelian fundamental group brings us to a contradiction.


For the sake of completeness, here is a proof of the last assertion.

Lemma For a path connected pointed topological space $(Y,y)$ the set $[S^1,Y]$ of homotopy classes of unpointed maps $S^1\to Y$ are in bijection with the set of conjugacy classes in $\pi_1(Y,y)$.

Proof. There is an evident map $\pi_1(Y,y)\to [S^1, Y]$. It is surjective because any unpointed map $f:S^1\to Y$ is homotopic to its conjugation by a path from $f(p)$ to $y$. Next, suppose that $f_0,f_1:S^1\to Y$ are pointed maps and $F:S^1\times[0,1]\to Y$ is an unpointed homotopy between them. The restriction $F|_{p\times[0,1]}$ induces a pointed map $g:S^1\to Y$ because $F((p,0))=f_0(p)=y=f_1(p)=F((p,1))$. We then have the equality $[f_0]=[g]^{-1}[f_1][g]$ in $\pi_1(Y,y)$. Indeed, a pointed homotopy between loops representing the sides of this equality is given by $$G(\alpha, t)=\begin{cases}g(3\alpha\cdot t),0\leq \alpha<\frac{1}{3}\\ F(3\alpha-1,t), \frac{1}{3}\leq\alpha <\frac{2}{3}\\ g(3(1-\alpha)\cdot t),\frac{2}{3}\leq\alpha\leq 1\end{cases}$$ (This formula defines a map $G:[0,1]\times [0,1]\to Y$ that factors through $S^1\times[0,1]$: the latter is the desired homotopy). Thus, any two elements of the fundamental group that can be represented by homotopic maps are conjugate, the converse is evident so the lemma is proven.

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    $\begingroup$ Unpointed homotopy classes $[S^1,Y]$ are given by the set of conjugacy classes in $\pi_1(Y)$, not $H_1(Y)$. (The former is the orbits of the conjugacy action of $\pi_1$ on itself in the category of sets, the latter in the category of groups (i.e. the abelianisation.) $\endgroup$ Nov 29, 2019 at 6:45
  • $\begingroup$ They always differ for nonabelian $\pi_1$, but for an easy example just think of a perfect group. $\endgroup$ Nov 29, 2019 at 6:45
  • $\begingroup$ It's an interesting fact that the fundamental and homotopy groups are dependant on having a base point, See also this discussion:mathoverflow.net/questions/40945 $\endgroup$ Nov 29, 2019 at 15:31
  • $\begingroup$ The use of groupoids leads to "higher homotopy groupoids" which are usually "nonabelian" but to get that needs, in my view, more structure on a space than just a base point, e.g. a filtered space, or n-cube of pointed spaces. $\endgroup$ Nov 29, 2019 at 15:39
  • $\begingroup$ @AchimKrause Yes, you are absolutely right. Thanks for the correction! $\endgroup$
    – SashaP
    Nov 29, 2019 at 16:45
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There is a more topological way. If you assume that $X$ admits a universal covering $\tilde X$ (so $X$ is path connected and semilocally simply connected, I believe) then the $G=\pi_1(X)$ is realized by the deck transformations, i.e. the self-homeomorphisms of $X$ that preserve the fibers of $p:\tilde X \to X$.

I am a bit worried, however, that my answer simply sweeps basepoints under the rug. They are certainly used in the construction I know of the universal cover $\tilde X$, but we can forget about them afterwards and simply use the covering map $p$.

**Edit: ** I still think this works, but I'm not sure if this construction is functorial in the sense you need.

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    $\begingroup$ I don't think you can make this construction functorial without picking base points. (And, indeed, the other answer proves that you can't.) $\endgroup$
    – HJRW
    Nov 30, 2019 at 12:41
  • $\begingroup$ See also the discussion at this question for arguments that you cannot make the universal cover construction functorial without choosing basepoints. $\endgroup$ Nov 30, 2019 at 14:42

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