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I did exercise 19 in Hatcher on page 132 and I was wondering if anyone could tell me if this is right:

19. Compute the homology groups of the subspace of $I \times I$ consisting of the four boundary edges plus all points in the interior whose first coordinate is rational.

My solution:

(i) The homology group of homotopy equivalent spaces is the same.

(ii) if $A \subset X$ is contractible then $X$ is homotopy equivalent to $X/A$

So by (i) and (ii) I can quotient the space by the top boundary union the bottom boundary union any one of the vertical lines. Doing this yields a wedge of circles $ X/A = \vee_i S^1$ where $i \in \mathbb{Q} \cap [0,1] - \{ * \}$.

Let $*$ denote the point where all the $S^1$ are attached to each other. Then $(S^1, \{*\})$ is a good pair i.e. there exists a neighbourhood of $\{*\}$ that deformation retracts onto $*$ (in fact every neighbourhood does).

(iii) if $(X,x_0)$ and $(Y,y_0)$ are good pairs then $H_k(X \vee Y) = H_k(X) \oplus H_k(Y)$

Using (iii), $H_k(X/A) = \oplus_i H_k(S^1)$ where $i \in \mathbb{Q} \cap [0,1] - \{ * \}$.

Now:

$k=0$: $H_0(X) = \mathbb{Z}$ because $X$ is path connected.

$k=1$: $H_1(X) = H_1(X/A) = \oplus_i H_1(S^1) = \oplus_i \mathbb{Z}$

$k>1$: $H_k(X) = \oplus_i H_k(S^1) = \oplus_i \{ 0 \}$ because $S^1 = e^0 \cup e^n$ i.e. the union of a zero cell and an $n$ cell. But in a cell complex $C$ of finite dimension $n$, $H_k(C) = \{ 0 \}$ for $k > n$.

Many thanks for your help!

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    @Matt: Continuous bijections are not homeomorphisms in general- the inverse would also need to be continuous. Nor are they homotopy equivalences. Consider the obvious map from the half open unit interval to the unit circle, for example.2011-05-18

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