# Cannonical group isomorpshim for fundamental group in a path connected space.

My question is related to this one which tells us that fundamental group $$\pi_{1}(X,x_0)$$ is abelian $$\textit{iff}$$ for every pair $$\alpha$$ and $$\beta$$ of paths from $$x_0$$ to $$x_1$$, we have the same group isomorphism.

let $$[I,X]$$ denote the set of homotopy classes of maps of $$I$$ into $$X$$, where $$I=[0,1]$$. If $$X$$ is path connected, we can easily prove that $$[I,X]$$ has only one element, which mean that every path in $$X$$ is homotopic to each other (even a loop at $$x_0 \in X$$ is homotopic to path $$p:x_0 \rightarrow x_1$$).

using this fact, $$\hat{\alpha}([f]) = [\bar{\alpha}]*[f]*[\alpha] = [\bar{\beta}]*[f]*[\beta]=\hat{\beta}([f])$$, wehre $$\alpha$$ and $$\beta$$ denote paths in path connected space $$X$$ ($$\alpha \simeq \beta \Rightarrow [\alpha]= [\beta]$$ ). So, it seems that there is no need for the fundamental group $$\pi_{1}(X,x_0)$$ to be abelian.

Also, for the proof that $$\pi_{1}(X,x_0)$$ is abelian, if you take 2 loops at $$x_0$$, then $$[f]*[g]=[g]*[f]$$ by virtue of being two paths in path-connected space $$X$$.

Am I wrong here? Where am I wrong?

• You are confusing free homotopies with base-point preserving homotopies. – Angina Seng Dec 1 '18 at 6:48
• I read this, but I am not sure how this concept of free homotopies and base-point preserving homotopies answer my question. One of the points that I still can't find any fault is that if $\alpha, \beta : x_0 \rightarrow x_1$, then $\alpha \simeq \beta$(since $X$ is path connected) which imply that $[\alpha] = [\beta]$ and hence, $\hat{\alpha}= \hat{\beta}$ – MUH Dec 1 '18 at 8:29
• @LordSharktheUnknown Also, if the free homotopies hold true, and given the base-point maps, (for e.g. here $\hat{\alpha},\hat{\beta}$, which are base-point maps at $x_1$ or $\textit{loops}$ at $x_1$), doesn't it imply $\hat{\alpha}=\hat{\beta}$ – MUH Dec 1 '18 at 8:40

Your conclusion that if $$\alpha, \beta : x_0 \rightarrow x_1$$, then $$\alpha \simeq_p \beta$$(since $$X$$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$$p:x_0 \rightarrow x_1$$ and $$q:y_0 \rightarrow y_1$$ are any two paths in path-connected space $$X$$. Then $$p \simeq e_{x_0}$$ and $$q \simeq e_{y_0}$$ and since, $$e_{x_0} \simeq e_{y_{0}} \Rightarrow p \simeq q$$. But this doesn't mean that $$p \simeq_p q$$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $$t \in I$$, which clearly doesn't get satisfied in the proof of homotopy ($$\textit{free homotopy}$$).