# Hatcher Universal Covering Space Construction - Basis

Below is an excerpt from Hatcher's Algebraic Topology. He is constructing a universal cover for a path-connected, locally path-connected, and semilocally simply-connected space $$X$$:

I don't understand why it follows that $$\mathcal{U}$$ is a basis for $$X$$. Is this a general fact that if a collection of open sets claimed to be a basis has the property that every basis element contains another basis element, then the collection is actually a basis? This doesn't sound right.

• Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book. Dec 9, 2018 at 20:41
• I haven't seen this before. The condition I'm familiar with is that given any open set $U \subset X$, for all $x \in U$, there is a basis element $V$ such that $x \in V \subset U$. Is that fact related to this condition? Dec 9, 2018 at 20:43
• That is what’s underlined in red. Dec 9, 2018 at 20:57

A basis for a topology $$\mathcal{T}$$ on a space $$X$$ is a subset $$\mathcal{B} \subset \mathcal{T}$$ such that for each $$U \in \mathcal{T}$$ and each $$x \in U$$ there exists $$B \in \mathcal{B}$$ such that $$x \in B \subset U$$.

A space $$X$$ is defined to be locally path connected if it has a basis consisting of path connected open sets (in other word, if the set $$\mathcal{P}$$ of path connected open sets forms a basis for $$X$$).

Hatcher shows that in a locally path connected semilocally simply-connected space $$X$$ the subset $$\mathcal{U} \subset \mathcal{P}$$ of all $$U \in \mathcal{P}$$ such that $$\pi_1(U) \to \pi_1(X)$$ is trivial also forms a basis for $$X$$. Note that this part does not use that $$X$$ is path connected.

• Ah, so I didn't have the right definition of locally path connectedness. I though locally path connectedness was defined to be the condition that for all $x \in X$, there is a neighbourhood of $x$ which is path connected. Dec 9, 2018 at 23:45
• This may be a silly question, but the definition of local path connectedness, as you said, is that for every $x \in X$ and for every neighbourhood $U$ of $x$, there is a path connected open set $V$ such that $x \in V \subset U$. In the definition of a basis, however, we have that for every open set $U$ and for every $x \in U$, there is a basis element $B$ such that $x \in B \subset U$. I don't see why local path connectedness means that we have a basis of path connected open sets. The quantifiers are in the wrong positions. Dec 9, 2018 at 23:48
• I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/… Dec 10, 2018 at 0:16
• If you require that each $x \in X$ has a path connected neighbourhood, then each path connected $X$ would be locally path connected. By the way, see also math.stackexchange.com/q/3000072. Dec 10, 2018 at 11:36
• Concerning you second comment: It is irrelevant in which order you arrange the quantifiers. Let us require (1) $\forall x \in X$ $\forall$ neighborhoods $U$ of $x$ $\dots$ Now consider any open $U$ and any $x \in U$. Then $x \in X$ and $U$ is a neighborhood of $x$, so (1) applies. Conversely let us require (2) $\forall$ open $U$ $\forall x \in U \dots$ Now consider any $x \in X$ and any neighborhood $U$ of $x$. Then $U$ is open and $x \in U$, so (2) applies. Dec 10, 2018 at 11:50

Let me iron all of this out.

Let $$U$$ be an open set of $$X$$ and $$x \in U$$. Since $$X$$ is locally path connected, $$X$$ has a basis $$\mathcal{B}$$ of path connected sets. Hence, we can find a $$B_1 \in \mathcal{B}$$ such that $$x \in B_1 \subset U$$. Since $$X$$ is semilocally simply connected, there is a an open set $$V$$ containing $$x$$ such that the induced inclusion $$\pi_1(V,x) \rightarrow \pi_1(X,x)$$ is trivial. Hence, we can find a $$B_2 \in \mathcal{B}$$ such that $$x \in B_2 \subset V$$ and furthermore we can find a $$B_3 \in \mathcal{B}$$ such that $$x \in B_3 \subset B_1 \cap B_2 \subset V$$. Then the composition of induced inclusions $$\pi_1(B_3,x) \rightarrow \pi_1(V,x) \rightarrow \pi_1(X,x)$$ is trivial and $$B_3 \subset U$$. Thus, what we have shown is that the subset of $$\mathcal{B}$$ consisting of elements $$W$$ such that the induced inclusion $$\pi_1(W) \rightarrow \pi_1(X)$$ is trivial is also a basis for $$X$$.

(Remember that since $$W$$ is path connected and the induced inclusion $$\pi_1(W,x) \rightarrow \pi_1(X,x)$$ is trivial for the choice of basepoint $$x$$ above, the induced inclusion is trivial for all other basepoints in $$W$$. Hence, it is acceptable to omit the basepoint and write that the induced inclusion $$\pi_1(W) \rightarrow \pi_1(X)$$ is trivial.)