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In discussing CW-complexes, how would I go about proving the following statement:

The homomorphism $i_* : H_q(X) \to H_q(X^*)$ is an isomorphism except possibly for $q = n$ and $q = n-1$.

Where $X^*$ is a space obtained by 'gluing on' $n$-cells onto space $X$, and $i_*$ is the homomorphism induced by the inclusion map?

Intuitively, I can't wrap my head around this result. For instance, if I take a circle and then I paste a $2$-cell onto the circle, then I start with a fundamental group of $\mathbb{Z}$ and by pasting the space onto it I end up with a trivial fundamental group (which doesn't contradict the theorem since in this case I am considering the map $$i_* : H_1(S^1) \to H_1(S^1 + U^2) = H_1(E^2)$$ (where $U^2$ is the open 2-dimensional disc, $E^2$ is the closed 2-dimensional disc and $S^1 + U^2$ is 'gluing' $U^2$ onto $S^1$). Here, $q = 1$ and $n = 2$, so $q = n-1$ and the theorem doesn't require the homomorphism to be an isomorphism so thats fine.

The problem is when I go up a dimension, and imagine pasting $U^3$ onto $S^1$. Now the inclusion homomorphism: $$i_* : H_1(S^1) \to H_1(S^1 + U^3)$$ is asserted to be an isomorphism. This means that $H_1(S^1 + U^3) = \mathbb{Z}$, and that pasting on a three-dimensional ball doesn't trivialise the fundamental group, whereas pasting on a two-dimensional ball does. But it seems like pasting on a three-dimensional ball should trivialise the fundamental group?

Kenny Wong
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QCD_IS_GOOD
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1 Answers1

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If you simply wish to prove this, then I suggest you write down the long exact sequence for the pair $(X^\star, X)$: $$ \dots \to H_{q+1}(X^\star, X) {\to} H_q(X) \overset{i_\star}{\to} H_q(X^\star) \to H_q(X^\star, X) \to \dots$$ Since $(X^\star, X)$ is a good pair, and since the quotient $X^\star / X$ is homeomorphic to a "bouquet" of $n$-spheres $\vee_{i=1}^k S_i^n$, we have $$H_q(X^\star, X) \cong \widetilde H_q (X^\star / X) \cong \widetilde H_q (\vee_{i=1}^k S_i^n) = \begin{cases} \mathbb Z^{\oplus k} & q = n \\ 0 & q \neq n\end{cases}$$ The result that $i_\star : H_q (X) \to H_q (X^\star)$ is an isomorphism for all $q \notin \{n-1, n \}$ then follows from the long exact sequence.

Presumably you would like an explanation of the intuition too, but I'll leave that to someone else...

Kenny Wong
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    For an intuition, when you glue $n$-cells, the cellular complex is unaffected except at $\Delta_n$. Since $H_p:=\ker \partial_p/\mathrm{im} \partial_{p+1}$, the $p$-th homology up to $p=n-2$ won't see anything that happened in $\Delta_n$. (This doesn't prove that the isomorphism is induced by inclusion, but this answer fills the bill). – Aloizio Macedo Mar 28 '17 at 18:35
  • What exactly is a good pair? I've seen the term used a few times but I don't have a definition of it? – QCD_IS_GOOD Mar 28 '17 at 18:47
  • $(X,A)$ is a good pair iff there exists an open neighbourhood $V$ of $A$ that deformation retracts onto $A$. This implies that there is a homotopy equivalence of pairs between $(V,A)$ and $(A,A)$, and from this, you use excision to prove that $H(X,A) \cong \widetilde H(X/A)$. If $X$ is the $n-1$ skeleton of a CW complex and $X^\star$ is the $n$-skeleton, then $(X^\star, X)$ is a good pair. – Kenny Wong Mar 28 '17 at 18:55