Fundamental group of the union of subspaces

Question

Let $X$ be a topological space, and let $Y$ be a subspace that is a union of nested open subspaces of $X$: $Y=\bigcup Y_n$, where $Y_0\subset Y_1\subset\dots$. Suppose the smallest subspace (hence all subspaces) contains the base point $x_0$. Show that if the inclusions $Y_i\subset Y_{i+1}$ induce the trivial maps on the fundamental groups, then the fundamental group of $Y$ is trivial and that in the event that $\pi_1(Y,x_0)$ is trivial, some homomorphism $\pi_1(Y_0,x_0)\to\pi_1(Y_i,x_0)$ is trivial.

The first part seems to be clear. Namely, suppose each inclusion $\iota: Y_i\subset Y_{i+1}$ induces the trivial map. Let $f$ be a loop in $Y_i$ centered at $x_0$. Then $\iota \circ f$ is homotopic to the constant loop at $x_0$ in $Y_{i+1}$ by assumption. But the loop $f$ is equal to $\iota \circ f$ (and it lies in $Y_i$), so $f$ is homotopic to the constant loop at $x_0$ in $Y_i$; therefore, $\pi_1(Y_i,x_0)=\{1\}$ for all $i$. So if $g$ is any loop in $Y$, then $g$ is a loop in some $Y_i$ for $i$ sufficiently large, and thus is a trivial element of $\pi_1(Y_i,x_0)$ and hence of $\pi_1(Y,x_0).$ Is this proof correct?

For the second part, I'm not sure how to proceed. I have only trivial thoughts. Take $[f]\in \pi_1(Y,x_0)$. It is trivial. $f$ lies in $Y_i$ for $i$ large. But how to relate it to $Y_0$?.. I guess I can show that $\pi_1(Y_0,x_0)\to\pi_1(Y_i,x_0)$ is injective for that chosen $i$, but that's not what's being asked.

Answer 1

Your proof is not correct : you cannot conclude that $\pi_1(Y_i, x_0)$ is trivial, because the homotopy taking $f$ to the constant loop may go outside $Y_i$.

But you don't need the triviality of this group to conclude afterwards.

And then later on you don't prove that $g$ is contained in some $Y_i$; you should at least mention in a few words why this is true.

The second part, however, is not true. Indeed, put $Y=\mathbb{R}^2$, and to describe the $Y_i$'s here's how they look : you take a strip like $(-1,1)\times (-1,\infty)$, and remove infinitely many points, specifically you remove $\{0\}\times\mathbb{N}$. That's your $Y_0$. Then $Y_i $, $i\geq 1$ is $Y_0$ together with the open ball of radius $i$ and centered at $0$.

Clearly the $Y_i$'s are open, they cover $Y=\mathbb{R}^2$, which has trivial fundamental group. However, $\pi_1(Y, 0)\to \pi_1(Y_i,0)$ is never the zero map.

To produce other counterexamples, try to think about how a supposed proof would go : you start from a loop $f$ in $Y_0$, it's homotopic to a constant loop in $Y$, the homotopy is entirely contained in a $Y_i$, so $f$ is sent to $0$ in $Y_i$. But wait, $i$ here depends on $f$ ! So to find a counterexample as I did, simply make the $i$ grow as $f$ changes : in my example this is what happens : the loops that go once around a missing point $(0,n)$ are homotopic to a constant map but not in $Y_n$.

Answer 2

In order to add to what has been said, here's an example of $Y_n$ such that each $Y_n$ is open, each has a non-trivial fundamental group and each inclusion induces a trivial group homomorphism:

So in each step you fill the hole and you add another one to the right. And you go to infinity. I know that borders are thick but you should think about them as open subsets of $\mathbb{R}^2$.

Each one of them is homotopically equivalent to $S^1$ and thus has $\mathbb{Z}$ as the fundamental group. Inclusions however induce trivial group homomorphism. This is equivalent to the fact that every loop in $Y_m$ is contractible in $Y_{m+1}$ (can you see why?).

Note that $Y$, which is equal to the union of all of them, is a contractible space.