0
$\begingroup$

I am studying for an exam and would appreciate if someone could spell me a proof on the following:

If $ X_0 \subset X_1 \subset \dots $ is a CW-complex with pushouts

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

where '$inc$' denotes the set-inclusion in all cases, then $$ H_n(<\varphi_i>_i) = \pm H_n(<\varphi_i'>_i) $$

What I do know, is that $$ H_n(\overline{<\phi_i>_i}),\,H_n(\overline{<\phi_i>_i}): H_n(\vee_{i \in I} (D^{n+1}/S^n)) \longrightarrow H_n(X_{n+1}/X_n) $$ differ by at most $-1$.

I have also tried toying around with the related diagrams and obvious relationship coming from the commutative diagrams in the category of topological spaces, but I haven't managed to prove this (and I have an exam coming up on Friday, so I don't have much time for this).
I have shown $$ H_n(X_n \rightarrow X_{n+1}/X_n) \circ H_n(<\varphi_i>_i) = \pm H_n(X_n \rightarrow X_{n+1}/X_n) \circ H_n(<\varphi_i'>_i) $$ (with the respective left maps being induced by the projection of $X_n$ into $X_n \in X_n+1/X_n$. However, I don't think this will help much, because this only shows $$H_n(<\varphi_i>_i) \mp H_n(<\varphi_i'>_i)(x) \in \ker(H_n(X_n \rightarrow X_{n+1}/X_n))\,,\; x \in H_n(\vee_{i \in I} (D^{n+1}/S^n))$$ (and I don't see how that is going to help).

I also tried playing around with the Mayer-Vietoris sequences for the pushout squares, but didn't come up with anything useful.

Looks like I'm going for a fail or bad grade as I'm way behind with the lecture notes, but help would still be greatly appreciated.

Note: At this point, I'm only starting to learn about cellular homology, so results from this cannot be used here.

EDIT: It's possible that I understood the question in a too general sense, and that the homework question was referring to a specific case (possibly a map that would be an answer to this question of mine ).
So maybe, this statement is wrong. I am leaving it open, because I would be curious about a counterexample in the general case, or a proof.

$\endgroup$
2
  • $\begingroup$ I'm learning cellular homology at that time like you . But I have some troubles with your notations and don't really understand what you're trying to do . Can you express it in another way ? $\endgroup$ Apr 19, 2017 at 15:20
  • $\begingroup$ @BangPhamKhoa I can't really help you, unless you are more specific... I can think of two things: The $<\varphi_i>_i$-part: Those are the maps coming from the universal property of the disjoint union/topological sum. The other: Putting the arrow-notation as the $H_n$-argument. But that simply means to take the homology of the obvious map (here: the projections precomposed with inclusion). $\endgroup$ Apr 19, 2017 at 16:46

1 Answer 1

0
$\begingroup$

It appears that at least some basic things about cellular homology are needed here.

In the lecture, we introduced two definitions for the cellular chain complex, namely, for $n \ge 0$:

$$ \begin{align} C^\text{cell, geo}_n(X) &:= \bigoplus_{e \in \Pi_0(X_n\setminus X_{n-1})} \mathbb{Z}\quad, \\ C^\text{cell}_n(X) &:= H_n(X_n\,, X_{n-1}) \end{align} $$

and both $0$ for for $n\lt0$, with a somewhat more contrived definition for the boundary homomorphisms in the geometric case based on the degrees of maps (for a fixed isomorphism-choice $D^n/S^{n-1}\cong S^n$) $$ S^n \overset{\varphi_i}\longrightarrow X_n/X_{n-1} \underset\cong{\overset {\bar{<\bar\psi_\nu>_\nu}} \longleftarrow} \vee_\nu D^n/S^{n-1} \overset{pr_l}\longrightarrow D^n/S^{n-1} \cong S^n $$ and indices $i$,$\,l$, where $\psi_l$ comes from an attaching diagram

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % \bigsqcup_{l \in L} S^{n-1} & \ra{\hspace{0.35cm} \hspace{0.35cm}} & X_{n-1} \\ \da{inc} & & \da{inc}\\ \bigsqcup_{l \in L} D^{n} & \ras{<\psi_l>_l} & X_{n}\\ \end{array} \\ $$

We then showed that these are chain-isomorphic, and the proof we used then implies the statement in this answer.

$\endgroup$

You must log in to answer this question.

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