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If $U$ is an open subset of $\mathbb{R}^2$, is it true that $H_2(U)=0$ and what can we say about $\pi_1(U)$? (For example, can we show that $\pi_1$ isn't perfect, e.g. $\pi_1(U)\neq 0\Rightarrow H_1(U)\neq 0$?)

For $\pi_1$, I'm aware that there is an answer here: fundamental groups of open subsets of the plane, but I'd rather not appeal to something like the classification of all noncompact surfaces to answer a question that should be basic.

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$\pi_1$ of a noncompact surface is free on some number of generators. See answers to this MathOverflow question. (I like the proof in the OP, but it's a bit overkill - John Stilwell's proof is great and elementary.) You're not going to be able to evade doing this to calculate the fundamental group of open subsets of the plane, because there's no reason those shouldn't be wild - how do you propose calculating the fundamental group of the complement of the Cantor set?

In any case, you can realize the free group on any number of generators $0 \leq n \leq |\Bbb N|$ by deleting the integers from $1,2, \dots, n$ from the plane.

In particular, $H_1 = 0 \implies \pi_1 = 0$.

The $n$th homology of any non-closed $n$-manifold is $0$ by a strong version of Poincare duality. Here's how I would prove it. By picking a proper Morse function $f: M \to \Bbb R$ with no $i \in \Bbb Z$ a critical value, you get an exhaustion by compact submanifolds $M_1 \subset M_2 \subset \dots \subset M$, where $M_i = f^{-1}[-i,i]$. Now use Mayer-Vietoris and classification of compact surfaces to calculate that $H_2(M_i) = 0$ for all $i$, and hence $H_2(M) = 0$. (By modifying the way you prove $H_n(M) = 0$ for compact $M$ with boundary, you get an alternate proof of the above fact for smooth manifolds. I don't think this is salvageable for non-smoothable manifolds.)

Why does every open $M \subset \Bbb R^2$ admit a proper Morse function? Well... admittedly I don't know an easy way to show this in this specific case, other than just proving it for all noncompact manifolds. Usually you find a closed emedding $i: M \hookrightarrow \Bbb R^k$ for some $k$ and pick the function $f(x) = \langle i(x), p\rangle$ for some $p$. I don't see any way to exploit the structure of open subsets of $\Bbb R^2$ to do this in a special way.

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  • I've a few doubts, it would be great if you explain...1 could it be possible that $M_i$ is a manifold with boundary?...2. can you explain a bit why $H_2(M_i)=0$ ?? – Anubhav Mukherjee Sep 23 '15 at 21:58
  • @Anubhav.K: I should have made some more assumptions on $M$: that it be noncompact and connected. Then it cannot have closed submanifolds without boundary, so $M_i$ must have boundary. That's now I conclude $H_2(M_i) = 0$, because such things deformation retract onto a wedge of circles. –  Sep 23 '15 at 22:04
  • but how did you conclude that $H_2(M_i) = 0$... can you please explain? I get an intuitive idea that $M_i$ has two boundaries, so it has to be deformation retract to some 1 dim cw complex...but can you provide me some concrete proof? – Anubhav Mukherjee Sep 23 '15 at 22:15
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    @Anubhav.K See here for the torus case. For general smooth manifolds with boundary, pick a triangulation, and "push in" n-simplices on the boundary to deformation retract onto a complex with fewer n-simplices. Eventually you have none left, and have deformation retracted onto an $(n-1)$-dimensional complex. I don't really want to write more details on that in the comments. –  Sep 23 '15 at 22:23