4
$\begingroup$

Assume $ X_{-1}:=\emptyset, X_0 \subset X_1 \subset \dots \subset X\;$ is a CW-complex with pushouts

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % \bigsqcup_{i \in I} S^{n-1} & \ra{\hspace{0.35cm} <\varphi_i>_i \hspace{0.35cm}} & X_{n-1} \\ \da{inc} & & \da{inc}\\ \bigsqcup_{i \in I} D^{n} & \ras{<\phi_i>_i} & X_{n}\\ \end{array}\quad,\;n \in \mathbb{N} \\ % $$

where '$inc$' denotes the set-inclusion in all cases.

How do I show that $$ H_n(X_n,\,X_{n-1}) \cong H_n(X_n/X_{n-1}, \ast) $$ is an isomorphism of groups? Note that $H_n$ denotes the $n$-th functor of singular homology. Also, the $\ast$ is shorthand for $\{X_{n-1}\}$, i.e. the class of $X_{n-1}$ under the canonical projection.

I tried using the LESes for the respective pairs, connecting them through the projection-induced morphisms. But I don't think that was enough...

I am studying for an exam and would appreciate if someone could help me with this, complete answers are appreciated. This isomorphy is part of a proof in the lecture notes, but the professor didn't bother to make a remark about it.

$\endgroup$
4
  • 3
    $\begingroup$ Doesn't this just follow from the fact that $(X_n , X_{n-1})$ is a good pair? $\endgroup$
    – Kenny Wong
    Commented Apr 19, 2017 at 19:47
  • $\begingroup$ I see it now, thank you! Do you want to post an answer (if not, I will do that)? $\endgroup$ Commented Apr 19, 2017 at 19:51
  • $\begingroup$ Feel free to post an answer yourself. :) $\endgroup$
    – Kenny Wong
    Commented Apr 19, 2017 at 19:51
  • $\begingroup$ Okay, thank you! Great help :) $\endgroup$ Commented Apr 19, 2017 at 19:52

1 Answer 1

4
$\begingroup$

As Kenny Wong pointed out, this follows, because $(X_n, X_{n-1})$ is a good pair. Stacking pushouts left to right, $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % \bigsqcup_{i \in I} S^{n-1} & \ra{\hspace{0.35cm} <\varphi_i>_i \hspace{0.35cm}} & X_{n-1} & \ra{\hspace{1.3cm}} & \ast \\ \da{inc} & & \da{inc} & & \da{}\\ \bigsqcup_{i \in I} D^{n} & \ras{<\phi_i>_i} & X_{n} & \ras{\hspace{0.9cm}} & X_n/X_{n-1}\\ \end{array}\quad,\;n \in \mathbb{N} \\ % $$ use that, because we have these pushouts, we can apply the following theorem from left to right:

Given a homology theory $h_*$(for instance, singular homology $H_*$) and a pushout of topological spaces $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % A & \ra{\hspace{0.9cm}} & X \\ \da{} & & \da{}\\ B & \ras{\hspace{0.9cm}} & Y\\ \end{array}\\ % $$ where $(B,\,A)$ is a good pair, then $(Y,X)$ is a good pair and homology induces isomorphisms $h_n(B,\,A) \cong h_n(Y,\,X)\,,\;n \in \mathbb{Z}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .