Here are some questions arising when I read Hatcher's Algebraic Topology P125. I will copy Hatcher's words and write down my questions in bold .$$$$ Let us find explicit cylcles representing generators of the infinite cyclic groups $H_n(D^n,\partial D^n)$ and $H_n(S^n)$. Replacing $(D^n,\partial D^n)$ by the equivalent pair $(\Delta ^n,\partial \Delta ^n)$,we will show by induction on n that the identity map:$i_n:\Delta^n \rightarrow \Delta^n $ , viewed as a singular n-simplex , is a cycle generating $H_n(D^n,\partial D^n)$. That it is a cycle is clear since we are considering relative homology. When n=0 it certainly represents a generator. For the induction step, let $\Lambda\subset \Delta^n$ be the union of all but one of the $(n-1)-\mathbb{dimensional}$ faces of $\Delta^n$. Then we claim there are isomorphisms$$H_n(D^n,\partial D^n)\xrightarrow{\approx}H_{n-1}(\partial\Delta^n,\Lambda)\xleftarrow{\approx}H_{n-1}(\Delta^{n-1},\partial \Delta^{n-1})$$ The first isomorphism is a boundary map in the long exact sequence of the triple $(\Delta^n,\partial\Delta^n,\Lambda)$, whose third terms$H_i(\Delta^n,\Lambda)$ are zero since $\Delta^n$ deformation retracts onto $\Lambda$, hence$(\Delta^n,\Lambda)\simeq (\Lambda,\Lambda) $. First question: What is the deformation retraction? The second isomorphism comes from the preceding proposition since we are dealing with good pairs and the inclusion $\Delta^{n-1}\hookrightarrow \partial \Delta^n$ as the face not contained \Lambda induces a homeomorphism of quotients $\Delta^{n-1}/\partial \Delta^{n-1}\approx \partial \Delta^n/\Lambda$. The induction step induces then follows since the cycle $i_n$ is sent under the first isomorphism to the cycle $\partial i_n$ which equals $\pm i_{n-1}$ in $C_{n-1}(\partial \Delta^n,\Lambda)$.Second question:why is it $\pm i_{n-1}$ rather than $i_{n-1}$? The exact sequence is as follows $$\cdots\rightarrow H_n(\partial \Delta^n,\Lambda)\rightarrow H_n(\Delta^n,\Lambda)\rightarrow H_n(\Delta^n,\partial \Delta^n)\rightarrow H_{n-1}(\partial\Delta^n,\Lambda)\rightarrow\cdots$$ To find a cycle generating $H_n(S^n)$ let us regard $S^n$ as two n-simplices $\Delta^n_1 $ and $\Delta^n_2$ with their boundaries identified in the obvious way,preserving the ordering of vertices.Third question:What is the meaning of the full sentence? How to regard $S^n$ as it? The difference $\Delta^n_1-\Delta^n_2$,viewed as a singular $n$-chain, is then a cycle ,and we claim it represents a generator of $H_n(S^n)$,assuming $n>0$ so that the latter group is infinite cyclic. To see this ,consider the isomorphisms$$\widetilde{H_n}(S^n)\xrightarrow{\approx} H_n(S^n,\Delta^n_2)\xleftarrow{\approx} H_n(\Delta^n_1,\partial \Delta^n_1)$$where the first isomorphism comes from the long exact sequence of the pair $(S^n,\Delta^n_2)$ and the second isomorphism is justified by passing to quotients as before. Under these isomorphisms the cycle $\Delta^n_1-\Delta^n_2$ in the first group corresponds to the cycle $\Delta^n_1$ in the third group , which represents a generator of this group as we have seen , so $\Delta^n_1-\Delta^n_2$ represents a generator of $H_n(S^n)$.
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I think I find cycles through Mayer-Vietoris Sequences,which is simpler to understand. Take $S^2$ for example.Step 1: build a tetrahedron whose faces are viewed as a singular 2-simplex. Let one of the faces map into the southern hemisphere. The face , a singular 2-simplex, has three edges. Let one of the edges map into the upper semicircle.Finally we have the generator of the reduced homology group $\widetilde{H_0}(S^0)$(one point substract the other point).Thus the four 2-simplex is a generator. – user12580 Oct 27 '15 at 13:20
1 Answers
1) A deformation retraction from $X$ to $A \subset X$ is a continuous map $F: X \times I \to X$ such that for all $x \in X$, $F(x,0) = x$, and $F(x,1) \in A$; and $F(a,1) = a$ for all $a \in A$. You can think of '$X$ deformation retracts to $A$' as saying that you can continuously shrink $X$ onto $A$. You can find lots of questions and answers on this site explaining deformation retraction.
2) In fact $\partial i_n$ is equal to $(-1)^ni_{n-1}$ in $C_{n-1}(\partial \Delta^n, \Lambda)$. The point is that Hatcher doesn't really care what sign it is.
3) Think of each of the simplices as hemispheres of $S^n$. Then you glue them together by identifying their boundaries. In the case $n=2$, the two simplices are triangles. Imagine gluing their edges together and then pulling their interiors away from each other. Then you end up with something that (topologically) resembles a sphere. Preserving the order of vertices is just a technicality to make the gluing behave properly.
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