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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 $X$:

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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.

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  • $\begingroup$ Yes, almost. This is one of the first theorems about bases that you see. It’s in any point set book. $\endgroup$
    – Randall
    Dec 9, 2018 at 20:41
  • $\begingroup$ 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? $\endgroup$
    – Frederic Chopin
    Dec 9, 2018 at 20:43
  • $\begingroup$ That is what’s underlined in red. $\endgroup$
    – Randall
    Dec 9, 2018 at 20:57

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Frederic Chopin
    Dec 9, 2018 at 23:45
  • $\begingroup$ 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. $\endgroup$
    – Frederic Chopin
    Dec 9, 2018 at 23:48
  • $\begingroup$ I found a source explaining the equivalence of the definitions: proofwiki.org/wiki/… $\endgroup$
    – Frederic Chopin
    Dec 10, 2018 at 0:16
  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Dec 10, 2018 at 11:36
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    $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Dec 10, 2018 at 11:50
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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.)

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