This question is motivated by Exercise 1.7 from Differential Forms in Algebraic Topology by Bott & Tu. The original question in the text concerns the de Rham cohomology of $\mathbb{R}^2$ with points $P$ and $Q$ deleted. When I computed $H^1$, I constructed two generators (specified by the integrals along contours around $P$ and $Q$), and used the fact that the space of one forms is two-dimensional to show that the cohomology group is $\mathbb{R}^2$.
If I let $X_k = \mathbb{R}^2 \setminus \{k~\text{points}\}$, then based on the computation, I feel tempting to conjecture that $H^1(X_k) = \mathbb{R}^k$, which coincides with my intuition that $X_k$ is homotopically a wedge sum of $k$ circles. On the other hand, the same argument of mine shows $H^1(X_k) = \mathbb{R}^2$ for all $k \ge 2$, but I have a hard time to understand/visualize this result.
If there's an error, could you point it out? If my computation is correct, could you offer your insight on the result? Thank you!
One way to think about this is de Rham cohomology gives you the free part of simplicial cohomology. For the case of $\mathbb{R}^{2}-\{p_1,\cdots, p_{n}\}$, you can show that the space is homotpically equivalent to $\bigvee_{i=1}^{n}\mathbb{S}^{1}_{i}$. Thus its first homology group should be $\mathbb{Z}^{n}$ and its de Rham cohomology group should be $\mathbb{R}^{n}$.
To visualize it, you might consider the case $\mathbb{R}^{2}-\{*\}$ first. In this case the de Rham cohomology is represented by a ray $(x,0):x>0$, whose Poincare dual is the angular form $\frac{1}{2\pi}d\theta$. If you have to remove $n$ points, then you are actually working with $n$ angular forms on the punctured plane. Each one of them is dual to a corresponding ray. Thus the picture is kind of "fuzzy".