Direct limit of fundamental homotopy groups of an increasing sequence of topological spaces

Question

I want to find $\pi_{1}(Y)$, the fundamental homotopy group of $Y$ which is the union of an increasing sequence of open topological spaces $Y_{n}$. I know $\pi_{1}(Y_{n})$ and $i_{n}: \pi_{1}(Y_{n}) \to \pi_{1}(Y_{n+1})$ the group homomorphisms induced by the injections $Y_{n} \to Y_{n+1}$, for every $n$. I was told to take the direct limit of groups of $\{\pi_{1}(Y_{n}), i_{n}\}$. I am new to the notion of direct limit and though I understood what is the direct limit of $\{\pi_{1}(Y_{n}), i_{n}\}$ I can't figure why exactly it gives us $\pi_{1}(Y)$?

Answer 1

First let's show that, given a loop $L$ with basepoint $x$ (that is, the image of some continuous map $f: S^1\rightarrow Y$ with $f((0, 1))=x$), there is some $n$ such that $L\subseteq Y_n$.

To do this, suppose $L=im(f)$ is a loop in $Y$. Then let $A_i=f^{-1}(Y_i)$; we have $S^1=\bigcup A_i$, and each $A_i$ is open (because each $Y_i$ is open and $f$ is continuous). But since $S^1$ is compact, this means that $\{A_i: i\in\mathbb{N}\}$ has a finite subcover - and since $A_i\subseteq A_{i+1}$, this means there is some $n$ such that $A_n=S^1$. But then $L\subseteq Y_n$, so we're done.

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By a similar argument, show that if two loops are homotopic in $Y$, then they are already homotopic in $Y_n$ for some $n$.

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Now the idea is: given an element $\alpha$ of $\pi_1(Y)$, pick a representative $L_\alpha$; then $L_\alpha\subseteq Y_n$ for some $n$, so we can associate to $\alpha$ an element of the direct limit of the homotopy groups of the $Y_i$s. Do you see how to show that this assignment is (1) well-defined (that is, independent of the choice of representative and of $n$) and (2) yields an isomorphism?