# Showing fundamental group of a particular space is trivial.

Let $$Z$$ be the subspace of $$\Bbb R^2$$ given by $$Z=\bigg(\{0\}\times[-1,1]\bigg)\bigcup\bigg\{(x,y):0 Next consider the quotient space, $$X=\frac{Z}{\{(0,0),(1,0)\}}.$$

I want to show, $$X$$ has trivial fundamental group. Here is my approach.

Since $$X$$ is path connected, it is enough to show, $$\pi_1(X,[(0,0)])$$ is trivial. So let $$f:[0,1]\to X$$ be a loop based at $$[(0,0)]$$. Since $$[0,1]$$ is compact and continuous image of a compact set is compact we have, $$f([0,1])$$ is a compact subset of $$X$$. Hence we have a $$1>\delta>0$$ such that, image of $$f$$ is contained in $$X_{\delta}$$, where $$X_{\delta}=\frac{\big(\{0\}\times[-1,1]\big)\bigcup\big\{(x,y):\delta\leq x\leq 1,y=\sin\big(\frac{\pi}{x}\big)\big\}}{\{(0,0),(1,0)\}}.$$ Now, $$X_{\delta}$$ is contractible space. So $$f:[0,1]\to X_{\delta}$$ is homotopically equivalent to constant loop. Now for the inclusion map $$i:X_{\delta}\hookrightarrow X$$ we have the induced inclusion map, $$i_*:\pi_1(X_{\delta},[(0,0)])\hookrightarrow \pi_1(X,[(0,0)])$$. Since $$\big[f:[0,1]\to X_{\delta}\big]\in \pi_1(X_{\delta},[(0,0)])$$ is a trivial element, so $$\big[f:[0,1]\to X\big]\in \pi_1(X,[(0,0)])$$ is also a trivial element.

My question is, am I right? If, not where is my fault. Another question is, is $$X_{\delta}$$ a contractible space? Thanks.

• @Arthur I have edited and added $\pi$. Thanks for pointing out. Oct 31, 2019 at 6:05

It's not true that any compact subset of $$X$$ is contained in an $$X_\delta$$, in fact $$X$$ is itself compact ($$Z$$ is compact as the closure of a bounded set in $$\mathbb R^2$$, therefore $$X$$ is compact as well, as a quotient of a compact space) : that was an unjustified claim and in fact it was wrong.
What will make $$X_\delta$$ appear is path connectedness, not compactness.
Then this is not a "fault", but you did not justify the claim that $$X_\delta$$ is contractible, although it should not be too hard to write down explicitly a contracting homotopy.
• But $X$ is path connected. Oct 31, 2019 at 13:07
• I guess that, $X_{\delta}$ homeomorphic to a space which looks like $\displaystyle\large\vdash$ and the later space is contractible. Am I right? Oct 31, 2019 at 13:27
• You are correct that it is path-connected, but $Z$ isn't, and a path $\gamma: [0,1]\to X$ has some $\epsilon<1$ such that the restriction to $[\epsilon, 1]$ lands in something that looks like $Z$, so cannot hit $X_\delta$'s, otherwise it would connect them with $0$ in $Z$. Now by standard arguments relating to $[0,1]$, you can make that $\epsilon=0$ Oct 31, 2019 at 13:31
• You are right about the shape of $X_\delta$, although I would write it as $\dashv$ Oct 31, 2019 at 13:32