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Let $X$ be a topological space. Then, the cone $CX$ over $X$ is always contractible. Hence, $CX \simeq \{p\}$, and $$ H_n(CX) = \begin{cases} \mathbb{Z}, &\quad n = 0,\\ 0, &\quad n \neq 0. \end{cases} $$ On the other hand, $X \cong X \times \{0\} \subset CX$ embeds as a subspace, whence $H_n(X) \subset H_n(CX)$, and we may form the relative homology $H_n(CX, X)$. But this seems flawed as if $X = \{0,1\}$, then $CX \simeq [0,1]$, so, $H_0(X) = \mathbb{Z}^2 \not \subset H_0(CX) = \mathbb{Z}$. I am quite blind to see my error.

The context is: I am trying to understand why $H_n(X/A)$ cannot be isomorphic to $H_n(X,A)$ for all topological pairs $(X,A)$. (I know that this holds only true if $A$ is a deformation retract of $X$, and in the above case, $X$ is a retract of $CX$ if and only if $X$ is contractible. In particular, if $X$ is not contractible, it is not a deformation retract of $CX$.)

Edit: Perhaps I am wrong, assuming that $H_n(X)$ contains $H_n(A)$ as a subgroup...

warzasch
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    $H_n(A)$ is in general not a subgroup of $H_n(X)$. We have a homomorphism induced by inclusion, but it doesn't have to be injective. For example, in the set $[0,1]$ the points $0,1$ belong to the same path component, and they become the boundary of a $1$-simplex, and so they are mapped to the same point. – Mark Dec 12 '22 at 10:12
  • Right! It seems sufficient for $H_n(A) < H_n(X)$ that $A$ is a retract of $X$. Is this also necessary? If $\iota_\star$ is injective, then there exists $g: H_n(X) \to H_n(A)$ with $g \circ \iota_\ast = \operatorname{id}$. The question then is: Does there exist a continuous map $r: X \to A$ such that $r_\ast = g$? – warzasch Dec 12 '22 at 10:39
  • I reckon that this is not true in general. If $X = A= {p}$, then there is no continuous map which realizes the multiplication by $2$ on $H_0(X)$. In fact, there is only one continuous map from $X$ to $X$, namely the identity $p \mapsto p$. – warzasch Dec 12 '22 at 10:59
  • Do you know what a cofibration is? – Paul Frost Dec 28 '22 at 14:17
  • I am afraid, no. – warzasch Dec 28 '22 at 17:05
  • Also note that even for very good inclusions, at $H_0$ you will have $H_0(X, A) \cong \tilde H_0(X/A)$, the right hand side being reduced homology. – ronno May 28 '23 at 10:24

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