$X$ is path connected, show that $X$ is not simply connected [duplicate]

Let $X=\{a,b,c,d\}$ and $T=\{\emptyset, X, \{a\},\{b\},\{a,b\},\{a,b,c\},\{a,b,d\}\}$. I have already shown that $X$ is path-connected. I now need to show that $X$ is not simply connected.

My attempt: I know that if $X$ is path-connected then $X$ is simply connected iff the fundamental group is trivial. So if I prove that the fundamental group is not trivial, then Im done. But how do I do so?

Solution 1:

Hint: let $Z = \{a,b,c\}$ with open sets $\{ \emptyset, \{a\},\{b\},\{a,b\}, \{a,b,c\}\}$. Prove that $H \colon Z \times I \to X$ defined via $H(z,t) = z$ for $t<1$ and $H(x,1) = c$ is continuous. What can you say about the homotopy type of $Z$? Can you write $X$ as the union of open subspaces with the homotopy type of $Z$? If so, what techniques give you information about a space from information about a covering?

Spoilers below,

The existence of $H$ shows that $Z$ is contractible. The open sets $A =X \setminus \{d\},B = X \setminus \{c\}$ are homeomorphic to $Z$, and they cover the whole space, with $A \cap B$ homeomorphic to a discrete space. Computing the (reduced) Mayer-Vietoris sequence gives $H_1(X) = \widetilde{H}_1(X) \simeq \widetilde{H}_0(A \cap B) \simeq \mathbb{Z} \neq 0$, and so $\pi_1(X) \neq 0$.

Another approach,

This finite space is associated to the poset $P = \{a,b,c,d\}$ with $a \leq c,d$ and $b \leq c,d$. The realization of the simplical complex associated to $P$ is a square, which is homotopy equivalent to a circle. Hence $X$ is weakly homotopy equivalent to $S^1$, and in particular it is not simply connected.