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Consider the following diagram $$\begin{array}{} \Bbb S^n \hookrightarrow \Bbb D^{n+1} \\ \big\downarrow \\ \Bbb R \Bbb P^{n} \end{array}$$

Show that pushout of the above diagram is homeomorphic to $\Bbb R \Bbb P^{n+1}.$

I don't understand how to proceed. I think the map between $\Bbb S^n$ and $\Bbb R \Bbb P^{n}$ is obtained by the projection map from $\Bbb S^{n}$ to $\Bbb R \Bbb P^{n}.$ But what map is taken between $\Bbb S^n$ and $\Bbb D^{n+1}\ $? Finally how to determine the pushout of the diagram? Any help in this regard will be appreciated.

Thanks in advance.

Tyrone
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Anil Bagchi.
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1 Answers1

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$\mathbb RP^k$ can be represented as the quotient space $S^k/\sim \phantom{.}$ where $\sim$ identifies antipodal points. Let $p : S^k \to \mathbb RP^k$ denote the quotient map. Let $S^k_+$ denote the closed upper hemisphere of $S^k$ which is canonicallly homeomorphic to $D^k$ via $h : S^k_+ \to D^k, h(x_1,\dots,x_{k+1}) = (x_1,\dots,x_k)$. The map $p' : D^k \to \mathbb RP^k, p'(\xi) = p(h^{-1}(\xi))$, is a continuous surjection (note that each equivalence class with respect to $\sim$ has a representative in $S^k_+$). Since $D^k$ is compact and $\mathbb RP^k$ is Hausdorff, $p'$ is closed map and hence a quotient map. It identifies antipodal points in the boundary $S^{k-1}$ of $D^k$ and embeds the open ball $\mathring D^k$ as an open subset into $\mathbb RP^k$.

So what is the pushout of your diagram? It is the adjunction space $\mathbb RP^n \cup_p D^{n+1}$ which is obtained from the disjoint union $\mathbb RP^n \sqcup D^{n+1}$ by identifying $\xi \in S^n \subset D^{n+1}$ with $p(\xi) \in \mathbb RP^n$. Let $q : \mathbb RP^n \sqcup D^{n+1} \to \mathbb RP^n \cup_p D^{n+1}$ denote the quotient map and $q' : D^{n+1} \to \mathbb RP^n \cup_p D^{n+1}$ its restriction. It is a continuous surjection (note that each $\eta \in \mathbb RP^n$ has the form $p(\xi)$ with $\xi \in S^n$, thus $q(\eta) = q(\xi)$). Hence $q'$ is a quotient map. Clearly $q'$ is injective on $\mathring D^{n+1}$ (no identifications are made here) and $q'(\mathring D^{n+1}) \cap q'(S^n) = \emptyset$. On $S^n$ the map $q'$ identifies antipodal points and therefore $\mathbb RP^n \cup_p D^{n+1} \approx \mathbb RP^{n+1}$ by our above considerations.

Paul Frost
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  • I don't understand why is $q'$ a quotient map. I don't think restriction of a quotient map is again a quotient map. Also it is clear that $q'$ is injective on $\mathring D^{n+1}$ and for any $\xi \in \Bbb S^n$ we have $q'(-\xi) = [p(-\xi)] = [p(\xi)] = q (\xi).$ So if I assume $q'$ to be a quotient map then the pushout is homeomorphic to $S = D^{n+1}/x \sim -x,$ $x \in \Bbb S^n.$ So I think we need to argue as to why $S \approx \Bbb R \Bbb P^{n+1}.$ What I only familiar with is that $\Bbb R \Bbb P^{n+1} \approx \Bbb S^{n+1}/C_2,$ where $C_2$ is the cyclic group of order $2.$ – Anil Bagchi. Apr 19 '21 at 20:28
  • $q'$ is a closed map because $D^{n+1}$ is compact and $\mathbb RP^n \cup_p D^{n+1}$ is Hausdorff. Closed maps are quotient maps. In the first part of my answer it is shown that $p'$ is a quotient map. But both maps $p', q'$ make the same identifications in $D^{n+1}$. – Paul Frost Apr 19 '21 at 21:52
  • Why are you saying that $\Bbb R \Bbb P^n \cup_p \Bbb D^{n+1}$ is Hausdorff? What I know is that If $X$ is regular and $A \subseteq X$ is closed then $X / A$ is Hausdorff. It is clear that $\Bbb R \Bbb P^n \sqcup \Bbb S^n$ is regular (in fact normal since both the spaces are compact and Hausdorff). But what is $A$ here? Another way of saying that a quotient space is Hausdorff if it is the collection of orbits under some continuous group action $:$ If $X$ is a compact and Hausdorff topological space and $G$ is a compact topological group acting continuously on $X$ then $X/G$ is Hausdorff. – Anil Bagchi. Apr 20 '21 at 02:58
  • Apply this to the quotient map $q$. The only nontrivial part in showing that $R$ is closed is that $C = {x, [y]) \in D^{n+1} \times \mathbb RP^n \mid q(x) = q([y]) }$ is closed. Note that $C = { (x,[x]) \mid x \in S^n} \cup { (-x,[x]) \mid x \in S^n}$ and show that both parts are closed. – Paul Frost Apr 20 '21 at 13:28
  • Aren't the sets ${(x,[x])\ |\ x \in \Bbb S^n}$ and ${(-x,[x])\ |\ x \in \Bbb S^n}$ same? Because $x \in \Bbb S^n \iff -x \in \Bbb S^n$ and $[x] = [-x],$ for every $x \in \Bbb S^n$ since $\Bbb R \Bbb P^n$ is obtained from $\Bbb S^n$ by identifying it's antipodal points (diametrically opposite points). I also don't understand as to how $C$ can be written as the union of the above two sets. – Anil Bagchi. Apr 20 '21 at 15:04
  • Let $X = \Bbb RP^n \sqcup D^{n+1}.$ Let $S = {(x,[x])\ |\ x \in S^n }.$ Then $S$ is nothing but the image of $S^n \times S^n$ in $X \times X$ under the continuous map $\text {id} \times p,$ where $p : S^n \longrightarrow \Bbb R P^n$ is the quotient map. Since $S^n \times S^n$ is compact and $X \times X$ is Hausdorff (since $X$ is Hausdorff) it follows that $S$ is going to be closed in $X \times X$ as continuous image of a compact set is compact and compact subspace of a Hausdorff space is closed. Am I right? – Anil Bagchi. Apr 21 '21 at 17:01
  • Question 1: You are right, the sets are the same. I should have seen that ;-) In fact it is obvious from the definition of $C$ that $C = { (x,[x]) \mid x \in S^n}$ because $q(x) = [x]$ and $q([y] = [y]$, thus $[y] = [x]$. Question 2: Yes, your argument is correct. $C$ is the image of the diagonal in $S^n \times S^n$. – Paul Frost Apr 21 '21 at 17:33
  • You should now answer your question https://math.stackexchange.com/q/4111166 . – Paul Frost Apr 21 '21 at 17:37