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Let $X$ be a topological space and $CX$ be its corresponding cone. Suppose $g$ is a $n-$ cycle of $X$, I want to show that $g$ is a boundary in $CX$.

According to the naturality of the cone construction, I can find a map $Cg: \Delta^{n+1} \to CX$, intuitively the boundary of $Cg$ should be $g$. How to calculate the boundary of $Cg$ formally?

z.z
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  • The boundary of $Cg$ is certainly not $g$. Ignoring signs, it is $C(\partial g) + g$. This follows by just writing down what the boundary faces are defined to be for $C(\partial g)$ and $\partial Cg$, respectively. –  Feb 10 '19 at 07:10
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    If you have a map $g:\Delta^n \to X$, then $Cg: C\Delta^n \to CX$. I think you should think about the identification $C\Delta^n \cong \Delta^{n+1}$, maybe starting with small dimensions to get the idea. – Justin Young Feb 10 '19 at 15:24
  • @MikeMiller I am assuming $g$ is a cycle. Can you maybe write an answer about this formula? I spent a lot of time trying to work out the details, but failed. – z.z Feb 10 '19 at 15:47
  • A cycle is a sum of simplices. I guess it's possible that $n$ could be even and the sides could pair off, but almost all of the time in practice this is not the case: a cycle is going to be a sum of more than one simplex. (This is why I responded for simplices, not cycles.) For a cycle, you verify that the $C(\partial g)$ terms of the various simplices cancel out. –  Feb 10 '19 at 16:31
  • My cycle means that its boundary is zero. – z.z Feb 10 '19 at 17:07
  • Yes, a cycle is a sum of simplices whose boundary is zero. In particular it is a sum of simplices. It is rarely a simplex. –  Feb 10 '19 at 18:38
  • Sorry I am confused, wouldn't that imply $C\partial g$ zero? – z.z Feb 10 '19 at 18:44

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First, to set conventions: $$CX = I\times X/\{1\}\times X$$ $$X \cong \{0\}\times X \subseteq CX$$ $$\Delta^k = \{(t_o,\ldots,t_k) \mid \sum t_i = 1\}\subseteq I^{k+1}$$ We define the vertices $e_m \in \Delta^k$ by $$e_m = (t_0,\ldots,t_k), t_i = 0 \text{ for } i\not = m \text{ and } t_m = 1$$ Now, we identify subsimplices of $\Delta^k$ by listing the $e_m$ in some order, for example $[e_0,\ldots, \hat{e_i},\ldots, e_k]$ is the simplex identified with $\Delta^{k-1}$ by omitting the vertex $e_i$. Of course, omitting no vertices, $[e_0,\ldots, e_k] = \Delta^k$. To avoid excessive indexing, I will abuse notation and use the letters $e_i$ for the standard vertices of any dimensional simplex. It should be clear from the context which dimension we are in.

Now, we identify $C\Delta^k\cong \Delta^{k+1}$ via a map: $$(s,e_i) \mapsto se_0 + (1-s)e_{i+1}$$ In this way, the cone point corresponds to $e_0$. (This map is a homeomorphism by the standard compactness/Hausdorffness argument.)

Now, we define a natural transformation $C: S_*(X) \to S_{*+1}(CX)$ (singular chain functor) by sending a singular simplex $g:\Delta^k \to X$ to $Cg: \Delta^{k+1} \to CX$ (abuse of notation using the identification above). If you compute $\partial Cg$, then by definition you have $$\partial Cg = \sum_{i=0}^{k+1} (-1)^i (Cg)\vert_{[e_0,\ldots,\hat{e_i},\ldots, e_{k+1}]}$$ Now using the above identifications, it should be clear that $$(Cg)\vert_{[e_0,\ldots,\hat{e_i},\ldots, e_{k+1}]} = C(g\vert_{[e_0,\ldots,\hat{e_{i-1}},\ldots, e_{k}]})$$ for $i>0$, and, $$(Cg)\vert_{[\hat{e_0},\ldots, e_{k+1}]} = g$$ using the identifications above. Thus, $\partial Cg = g - C(\partial g)$, in other words, $C$ is a (natural) nullhomotopy of the map induced by the inclusion $X\to CX$ on singular chains. It follows immediately that this map sends cycles to boundaries.

Justin Young
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  • Thank you for your answer. By the way, I think it is way easier to understand if I think of cone as the formal convex combination of points. – z.z Feb 18 '19 at 03:16