If $X$ and $Y$ are path-connected topological spaces such that the fundamental group of $X$ is isomorphic to the fundamental group of $Y$, does it follow that the fundamental groupoid of $X$ is isomorphic or equivalent to the fundamental groupoid of $Y$?

# Fundamental group determines fundamental groupoid

algebraic-topologyfundamental-groups

#### Related Solutions

Your definition of $f$ is correct. (I will assume your definition of the map $\Pi\times_X \Pi\rightarrow X\times X$ is the intended one as this seems to me the only reasonable choice.) For notational convenience, given $p:[0, 1]\rightarrow X$, define $\tilde p$ to be the unique lift to $\Pi$ of $c_{p(0)}\times p:[0, 1]\rightarrow X\times X$ that starts at $e(p(0))$. In particular, we have $f[p]=\tilde p(1)$.

Now, given $([\gamma], [\eta])\in\tilde X\times_X\tilde X$ with representatives $\gamma, \eta:[0, 1]\rightarrow X$ (so in particular $\gamma(0)=\eta(1)$), we wish to show that $f[\gamma\bullet\eta]=m(f[\gamma], f[\eta])$, ie that $\widetilde{\gamma\bullet\eta}(1)=m(\tilde\gamma(1), \tilde\eta(1))$. Recall that $\widetilde{\gamma\bullet\eta}$ is the unique lift to $\Pi$ of $c_{\eta(0)}\times (\gamma\bullet\eta)$ that starts at $e(\eta(0))$, and note that $c_{\eta(0)}\times (\gamma\bullet\eta)=(c_{\eta(0)}\times\gamma)\bullet(c_{\eta(0)}\times\eta)$. Hence, for $\theta\in[0, 1]$, we have that $\widetilde{\gamma\bullet\eta}(\frac{\theta}{2})$ coincides with the unique lift to $\Pi$ of $c_{\eta(0)}\times\eta$ that starts at $e(\eta(0))$, ie with $\tilde\eta$. In particular, $\widetilde{\gamma\bullet\eta}(\frac{1}{2})=\tilde\eta(1)$, and hence, for $\theta\in[0, 1]$, we have that $\widetilde{\gamma\bullet\eta}(\frac{1+\theta}{2})$ coincides with the unique lift to $\Pi$ of $c_{\eta(0)}\times\gamma$ that starts at $\tilde\eta(1)$. In particular, $\widetilde{\gamma\bullet\eta}(1)$ is the endpoint of the unique lift to $\Pi$ of $c_{\eta(0)}\times\gamma$ that starts at $\tilde\eta(1)$.

Hence, by uniqueness of lifts, to show that $\widetilde{\gamma\bullet\eta}(1)=m(\tilde\gamma(1), \tilde\eta(1))$, it suffices to exhibit a lift to $\Pi$ of $c_{\eta(0)}\times\gamma$ that starts at $\tilde\eta(1)$ and ends at $m(\tilde\gamma(1), \tilde\eta(1))$. We claim $\tilde p:[0, 1]\rightarrow\Pi$, given by $\theta\mapsto m(\tilde\gamma(\theta), \tilde\eta(1))$, defines such a lift.

First note that $\tilde p$ is well-defined. To see this we must check that $s(\tilde\gamma(\theta))=t(\tilde\eta(1))$ for each $\theta\in[0, 1]$. But indeed, recall that $\tilde\gamma$ is a lift of the path $c_{\gamma(0)}\times\gamma$, so – for each $\theta$ – $\pi(\tilde\gamma(\theta))=(\gamma(0), \gamma(\theta))$, and in particular $s(\tilde\gamma(\theta))=\gamma(0)$. Likewise, $\tilde\eta$ is a lift of the path $c_{\eta(0)}\times\eta$, so $\pi(\tilde\eta(1))=(\eta(0), \eta(1))$, and in particular $t(\tilde\eta(1))=\eta(1)=\gamma(0)$, as desired.

Now we must show that $\tilde p(0)=\tilde\eta(1)$ and that $\tilde p(1)=m(\tilde\gamma(1), \tilde\eta(1))$. The latter condition is immediate, and to see the former note that, by definition of $\tilde\gamma$, we have $\tilde\gamma(0)=e(\gamma(0))=e(\eta(1))$. Hence $\tilde p(0)=m(\tilde\gamma(0), \tilde\eta(1))=m(e(\eta(1)), \tilde\eta(1))=\tilde\eta(1)$, as desired, where the last equality follows from the groupoid identity $m(e\times\text{id})=\text{id}$.

Finally, we need only show that $\pi\circ\tilde p=c_{\eta(0)}\times\gamma$, ie that $\tilde p$ really is a lift of $c_{\eta(0)}\times\gamma$. But indeed, for $\theta\in[0, 1]$, by definition of the map $\Pi\times_X\Pi\rightarrow X\times X$ and the fact that $m$ is a map over $X\times X$, we have $\pi(\tilde p(\theta))=\pi(m(\tilde\gamma(\theta), \tilde\eta(1)))=(s(\tilde\eta(1)), t(\tilde\gamma(\theta)))=(\eta(0), \gamma(\theta))=(c_{\eta(0)}\times\gamma)(\theta)$, as desired, where the second to last equality follows from the computations in paragraph 4. This concludes the proof.

No. Consider two spheres $S^n$ and $S^m$ with $n,m>1$. Then $\pi_1(S^n)=\pi_1(S^m)=0$ while $S^n$ is not homotopy equivalent to $S^m$ whenever $n\neq m$ (which can be seen by simple calculation of homology groups).

For more sophisticated example consider the double comb space $X$, which has all homotopy groups trivial $\pi_n(X)=0$, homology groups trivial $H_n(X)=0$ and cohomology groups trivial $H^n(X)=0$, but it is not contractible (it is not homotopy equivalent to a point).

## Best Answer

They are equivalent, but do not need to be isomorphic, as pointed out by Mark Saving in the comments.

Let $x_0 \in X$ be the basepoint. Since $X$ is path-connected, there exists an equivalence of the delooping of $\pi_1(X,x_0)$ to the fundamental groupoid $\Pi_1(X)$. The delooping of $\pi_1(X,x_0)$ can be seen as the full subgroupoid of $\Pi_1(X)$ with the only object $x_0$. This is sometimes denoted as $\Pi_1(X,x_0)$.

An equivalence $F:\Pi_1(X) \to \Pi_1(X,x_0)$ corresponds to the choice of a morphisms $\eta_x \in \mathrm{Hom}_{\Pi_1(X)}(x_0, x)$ for each $x \in X$.