Show that $\mathbb{R}P^5$ and $ \mathbb{R}P^4 \vee S^5$ have the same fundamental group and homology groups. Show they are not homotopy equivalent, because they have different higher homotopy groups.
Hints:
1- $S^n$ is a covering space of $\mathbb{R}P^n$.
2- For the last part of Problem 6, show that $\pi_4(\mathbb{R}P^5)=0$,but $\pi_4(\mathbb{R}P^4 \vee S^5)$ is non-zero. You can do this by arguing that $S^5 \vee S^4 \vee S^5$ is a covering space of $\mathbb{R}P^4 \vee S^5$ and then show $S^4$ is a retract of $S^5 \vee S^4 \vee S^5.$ You may assume $\pi_4(S^4)$ is isomorphic to the integers.
My questions are:
1- How can I prove that they have the same fundamental group?
2- How can I prove that they have the same homology group?
3- How can I argue that $S^5 \vee S^4 \vee S^5$ is a covering space of $\mathbb{R}P^4 \vee S^5$?
4- How can I show $S^4$ is a retract of $S^5 \vee S^4 \vee S^5$?
Could anyone help me in answering these questions please?
Best Answer
This answer might be overkill considering that the body of OP's question only asks for answers to 4 subproblems, but for completeness it might be worth it to spell out the solution to the whole question in the subject. (I recommend you first try to work out for yourself the hints in Thomas Rot's comment and then put the results together, because this answer just does that.)
This answer assumes that the following result is available, because it is very useful for this problem: if $\rho\colon E\to B$ is a covering map of connected spaces then the projection map induces an inclusion $\rho_*\colon \pi_1 E \hookrightarrow \pi_1 B$ whose image has index $n$, and $\pi_i E \cong \pi_i B$ for $i > 1$. (This can be proven using only material from Hatcher's AT Chapter 1, in particular see Propositions 1.31, 1.32 and 4.1. It also follows immediately from the long exact sequence of homotopy groups for a fibration. )
Answers to your questions (based on the hints):
1) The quotient map $S^n \to \mathbb{R}P^n$ is a 2-sheeted covering space for every $n$, so it follows from the assumed result that $\pi_1(\mathbb{R}P^n) \cong \mathbb{Z}/2$ for $n > 1$ since $S^n$ is simply connected in this range. Then by the Seifert-van Kampen theorem $$\pi_1(\mathbb{R}P^4 \vee S^5) \cong \pi_1(\mathbb{R}P^4) * \pi_1(S^5) \cong \mathbb{Z}/2 \cong \pi_1(\mathbb{R}P^5) $$
2) The homology of real projective space $\mathbb{R}P^n$ consists of $\mathbb{Z}$ in degree $0$, $\mathbb{Z}/2$ in every odd degree $<n$, $0$ in all even degrees, and if $n$ is odd then it is orientable so there is also a $\mathbb{Z}$ in degree $n$. Moreover, by a Mayer-Vietoris argument it follows that $H_i(X\vee Y) \cong H_i(X)\oplus H_i(Y)$ for all pointed spaces $X$ and $Y$ and all degrees $i > 0$. Since $\mathbb{R}P^4 \vee S^5$ is connected, we get the graded group isomorphisms
$$ H_\bullet(\mathbb{R}P^4 \vee S^5) \cong \mathbb{Z} \oplus \mathbb{Z}/2 \oplus 0 \oplus \mathbb{Z}/2 \oplus 0 \oplus \mathbb{Z} \oplus 0 \oplus \dots \cong H_\bullet(\mathbb{R}P^5)$$
3) Note that the quotient map $S^4 \to \mathbb{R}P^4$ identifies the antipodal points in the sphere. Form the space $X$ by attaching one copy of $S^5$ to $S^4$ at the south pole $S$, and another copy of $S^5$ at the north pole $N$ (note that strictly speaking this is not the coproduct of pointed spaces $S^5\vee S^4\vee S^5$ because there are multiple attaching points, but they are homotopy equivalent). Then since the antipodal map identifies $S$ and $N$ we can construct a function $X \to \mathbb{R}P^4 \vee S^5$ by taking the quotient map on $S^4$ and have the attached copies of $S^5$ map to $S^5$ by the identity. Away from the wedge point in $\mathbb{R}P^4 \vee S^5$ it's clear that that every point is evenly covered by 2 sheets (use small charts). If $x_0$ is the wedge point, then since $\mathbb{R}P^4$ and $S^5$ are both manifolds we can choose open balls $U_1$ and $U_2$ containing $x_0$ from each space, and let $U = U_1 \cup U_2$. If $U_1$ is small enough then its inverse image is two open balls $\tilde{U}_1, \tilde{U}_1'$ in $S^4$, and the inverse image of $U_2$ will be two copies $\tilde{U}_2, \tilde{U}_2'$ of itself, one in each copy of $S^5$. The inverse image of $U$ is then the disjoint union of $\tilde{U}_1 \vee \tilde{U}_2$ and $\tilde{U}_1' \vee \tilde{U}_2'$.
4) Recall that a subspace $A\subset X$ is a retract if there is a continuous function $r \colon X \to A$ such that $r|_A = id_A$. Notice that any wedge-summand is a retract, since we can continuously define $A\vee B \to A$ by mapping $A$ identically to itself and $B$ to the wedge point; in particular $S^4$ is a retract of $S^5\vee S^4 \vee S^5$. Note that if $A \stackrel{i}{\hookrightarrow} X \stackrel{r}{\to} A$ is a retract and $\pi_k(A) \neq 0$ for some $k$, then $\pi_k(X)$ is also non-zero since functoriality of $\pi_k$ gives a factorization $id_{\pi_k A} = \pi_k(r) \circ \pi_k(i)$.
Putting it all together: 1) and 2) say that $\mathbb{R}P^4\vee S^5$ and $\mathbb{R}P^5$ have the same fundamental group and homology groups. However, we can show they are not homotopy equivalent by using 3) and 4) to show they have different $\pi_4$. First, since $S^5$ is a covering space of $\mathbb{R}P^5$ they have the same higher homotopy groups, and in particular $\pi_4\mathbb{R}P^5 \cong \pi_4 S^5 = 0$. On the other hand, since $E = S^5 \vee S^4 \vee S^5$ is (homotopy equivalent to) a 2-sheeted covering of $\mathbb{R}P^4\vee S^5$ it suffices to compute $\pi_4 E$. Since we have a retraction $S^4 \stackrel{i}{\hookrightarrow} E \stackrel{r}{\to} S^4$ and $\pi_4 S^4 \cong \mathbb{Z}$ if follows from point 4) that $\pi_4 E$ is non-zero. Therefore $\pi_4(\mathbb{R}P^4\vee S^5) \not\cong \pi_4 \mathbb{R}P^5$.