Is there a topological space $X$ whose fundamental group depends on the base point?

Question

Is there a space $X$ such that $\pi_1(X, x) \ne \pi_1(X, y)$ where $x \ne y$? I have not been able to think of an example. Part of the reason is that I have not computed too many fundamental groups yet.

Answer 1

If you know any space $A$ and point $x\in A$, and any other space $B$ and point $y\in B$, such that $$\pi(A,x)\not\cong\pi(B,y),$$ we can let $X$ be the disjoint union $A\sqcup B$ (note that this makes $X$ disconnected). Because any continuous loop $\gamma:[0,1]\to X$ will have a connected image, any continuous loop must go entirely into either the "$A$ part" of $X$ or the "$B$ part" of $X$; thus, if I require my loop to start at $x\in A$, it will never "know" that a copy of $B$ was attached, and we will have $$\pi(X,x)\cong \pi(A,x).$$ Similarly, we will have $$\pi(X,y)\cong \pi(B,y).$$ Thus, we will have a space $X$ with points $x,y\in X$ for which $\pi(X,x)\not\cong\pi(X,y)$.

Here is a specific example: let $X$ be the disjoint union of a circle and a point, i.e. $$X=\mathbb{S}^1\sqcup\{y\}.$$ Then for any $x\in\mathbb{S}^1$, we have that $\pi(X,x)\cong\mathbb{Z}$, whereas $\pi(X,y)$ is the trivial group.

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Note that this is the sort of thing that will have to go wrong. It is a fact that if $X$ is path-connected, i.e. for any two points of $X$, there is a path starting at one and ending at the other (note that this is a stronger requirement than being connected), then $\pi(X,x)\cong\pi(X,y)$ for any $x,y\in X$.