Wrong proof in May's Algebraic Topology? Here is the link of the book: https://www.maths.ed.ac.uk/~v1ranick/papers/maybook.pdf
In the proposition in p.169 of the book (not the pdf), it is written that $H_n(\mathring{M},\mathring{M}-\mathring{U})\cong H_n(M,M-y)$, where $U$ is a coordinate chart at a boundary point and $y\in \mathring{U}$. But if we take $M=D^2$ and $U$ a small chart near $\partial D^2$, then it seems the two groups are not isomorphic. Am I missing something?
 A: Your doubts are justified. May writes

Consider a coordinate chart $U$ of a point $x ∈ ∂M$. If $\dim M = n$, then $U$ is homeomorphic to an open half-disk in $\mathbb H^n$. Let $V = ∂U = U ∩ ∂M$ and let
$y ∈ \mathring U = U \setminus V$.

So let us restrict to the special case $M = \mathbb H^n$ (you  can also take $M = D^n$ if you want a compact example). Let $D \subset \mathbb R^n$ be the closed ball with radius $1$ and center $x$, let $B = D \cap \mathbb H^n$ be the associated closed half ball and let $U$ be its interior in $\mathbb H^n$ . Clearly $B$ strongly deformation retracts to $B_0 = \partial B \cap \mathbb H^n$ via a homotopy $H : B \times I \to B$ which is stationary on $B_0$ such that $H_0 = id$, $H_1$ is a retraction to $B_0$ and $H_t(U) \cap B_1 = \emptyset$ for $t > 0$, where $B_1 = B \cap \partial \mathbb H^n$. Extend this by the stationary homotopy on $\mathbb H^n \setminus U$ to a homotopy $R : \mathbb H^n \times I \to \mathbb H^n$. By construction it restricts to a homotopy on $\mathring{\mathbb H}^n$ which shows that $\mathring{\mathbb H}^n$ strongly deformation retracts to $\mathring{\mathbb H}^n  \setminus \mathring U$. Therefore all $H_m(\mathring{\mathbb H}^n , \mathring{\mathbb H}^n  \setminus \mathring U)  = 0$.
On the other hand, exision shows that $H_m(\mathbb H^n, \mathbb H^n \setminus y) \approx H_m(C, C \setminus y)$ for some small closed ball $C \subset \mathring U$ centered at $y$. It is well-known that for $m = n$ we obtain $H_n(C, C \setminus y) \approx R$.
