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 $\text{For}:\text{ppTop}\to \text{pTop}$.



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> > 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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Can we define fundamental groups functorially for non-pointed path connected topological spaces?
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.




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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.


