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Hello I was studying cohomology from this notes https://www2.warwick.ac.uk/fac/sci/maths/people/staff/vincent/cohomology.pdf and on page 16 it gives the definition of an excision map:

For a space $X$ and $U \subseteq A\subseteq X$ the map of pairs $(X-U,A-U)\hookrightarrow(X,A)$ is an excision if the induced morphism $H^n(X-U,A-U)\rightarrow H^(X,A)$ is an isomorphim for all n.

Ten it claims for $X=\mathbb{S}^n$ , $A=D^-=\{x \in \mathbb{S}^n : x_n \leq0\}$), $U=\overset{\circ}{D^-}$ the map

\begin{align}(D^+, \mathbb{S}^{n-1}) \overset{i}{\hookrightarrow} (\mathbb{S}^n,D^-) \end{align} is an excision.

Now by the universal coefficients theorem I reduced the problem to showing that the induced map on the homology is an isomorphism.

Now, this is not trivial since $\overline{U} \nsubseteq \overset{\circ}{D^-}$ and thus you cannot use the excision theorem.

My idea is that this should be true because for a slightly smaller $U$, call it $\widetilde{U}=\{x \in \mathbb{S}^n : x_n < - \varepsilon\}$ the theorem is true.

The trick would be to show that there is a Strong Deformation Retract \begin{align} (X-U,A-U)\overset{j}{\hookrightarrow}(X-\widetilde{U},A-\widetilde{U}) \end{align}

Then you would have a conmutative square

$$ \require{AMScd} \begin{CD} (X-U,A-U) @>j>> (X-\widetilde{U},A-\widetilde{U})\\ @ViVV @VVkV \\ (X,A) @= (X,A)\\ \end{CD} $$

where all the arrows are inclusions of pairs except the lower one which is the identity. By the conmutativity of the diagram and the isomorphisms induced by $j,Id$ and $k$ (by classical excision, for $\overline{U} \subseteq \overset{\circ}{A}$), then $i$ would be an isomorphism.

Am I on the right track? Is $j$ indeed a Strong Deformation Retract ? I think that it obviously is but can't write the exact formula.

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To show that is an excision it suffices to prove that it induces isomorphisms in the homology. By a lemma in "Elements of algebraic topology" in Hatcher's book, a map of pairs $ f:(X,A) \rightarrow (Y,B)$ induces isomorphisms at the homology level if $f:X \rightarrow Y$ and $f|A \rightarrow B$ are both homotopic equivalences.

$\mathbb{S}^{n} -\overset{\circ}{D^-} = \{x \in \mathbb{S}^{n}\ : x \geq 0 \} $

$\mathbb{S}^{n} -\widetilde{U} = \{x \in \mathbb{S}^{n}\ : x \geq -\varepsilon \}$

$A-U = \mathbb{S}^{n-1}$

$A-\widetilde{U} = \{x \in \mathbb{S}^{n}\ : -\varepsilon \leq x \leq 0 \} $

Both

$\{x \in \mathbb{S}^{n}\ : x \geq 0 \} \hookrightarrow \{x \in \mathbb{S}^{n}\ : x \geq -\varepsilon \}$

and

$\mathbb{S}^{n-1} \hookrightarrow \{x \in \mathbb{S}^{n}\ : -\varepsilon \leq x \leq 0 \}$

induce strong deformations retract (contracting radially from $x=-\varepsilon$ to the ecuator)and thus by the lemma the inclusion induces isomorphisms at the homology level.

I strongly suspect this is actually an homotopy equivalence of pairs and the retraction would be contract radially from $x=-\varepsilon$ to the ecuator and leave $D^n_+$ unchanged

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