Every map $\mathbb{RP}^4 \to S^3$ induces the zero map on $\pi_4(\mathbb{RP}^4)$ I'm trying to show that every map $f : \mathbb{RP}^4 \to S^3$ induces the zero map on $\pi_4(\mathbb{RP}^4)$. Here's what I've tried:
Using cellular approximation, we can assume that $f$ is cellular, so $f$ factors through $\mathbb{RP}^4/\mathbb{RP}^2 \simeq \Sigma^2 (\mathbb{RP}^2)$. I thought that somehow I could use the facts that $S^3 \simeq \Sigma^2(S^1)$ and that every map $\mathbb{RP}^2 \to S^1$ is null-homotopic. But then I got stuck.
Another approach that I've tried is to show that $f$ factors through $\mathbb{RP}^5$ (which is equivalent to what I'm trying to prove, if I'm not mistaken), but I got stuck again.
I would really appreciate any hint.
 A: Let 
$$q:\mathbb{R}P^4\rightarrow S^4\cong \mathbb{R}P^4/\mathbb{R}P^3$$
be the map pinching to the top cell. Here are your hints: Show that $(1)$ every map $\mathbb{R}P^4\rightarrow S^3$ factors through $q$, $(2)$ $q$ induces the trivial map on all homotopy groups. Details follow.
$(1)$ Apply $[-,S^3]$ to the cofibration sequence 
$$S^3\xrightarrow{\gamma}\mathbb{R}P^3\xrightarrow{i}\mathbb{R}P^4\xrightarrow{q}S^4\rightarrow\dots$$
where $\gamma$ is the 2-sheeted universal covering and $i$ is the subcomplex inclusion. You get
$$\pi_3S^3\xleftarrow{\gamma^*}[\mathbb{R}P^3,S^3]\xleftarrow{i^*}[\mathbb{R}P^4,S^3]\xleftarrow{q^*}\pi_4S^3\leftarrow\dots$$
Now $\mathbb{R}P^3$ is an orientable 3-manifold, so by the Hopf degree theorem $[\mathbb{R}P^3,S^3]\cong \mathbb{Z}$ as a group, and is generated by the map $p:\mathbb{R}P^3\rightarrow S^3$ which pinches to the top cell. By studying homology you know that $p_*:H_3\mathbb{R}P^3\rightarrow H_3S^3$ is an isomorphism, and that $\gamma_*:H_3S^3\rightarrow H_3\mathbb{R}P^3$ is multiplication by $2$. The conclusion from this is that the composite $p\gamma=\gamma^*(p):S^3\rightarrow S^3$ is the degree $2$ map. From this you see that the left-most map in the Puppe sequence above is the injection $\gamma^*=\times 2:\mathbb{Z}\rightarrow\mathbb{Z}$. Thus you get exactness of 
$$0\leftarrow[\mathbb{R}P^4,S^3]\xleftarrow{q^*}\pi_4S^3\leftarrow\dots$$
and conclude $(1)$.
$(2)$ Let $\rho:S^4\rightarrow \mathbb{R}P^4$ be the universal cover. Recall that $\pi_1\mathbb{R}P^4\cong\mathbb{Z}_2$, and $\rho_*:\pi_kS^4\rightarrow\pi_k\mathbb{R}P^4$ is an isomorphism for $k\geq 2$.
By studying homology you find that the map induced by the composite $S^4\xrightarrow{\rho} \mathbb{R}P^4\xrightarrow{q} S^4$ is the zero homomorphism. This is true because it passes through $H_4\mathbb{R}P^4=0$, which is trivial because $\mathbb{R}P^4$ is non-orientable. The conclusion is that $q\rho:S^4\rightarrow S^4$ is null-homotopic.
By the above remarks we know that $\pi_4\mathbb{R}P^4$ is isomorphic to $\mathbb{Z}$ and is generated by $\rho$. Thus any element in $\pi_4\mathbb{R}P^4$ can be represented as a multiple $n\cdot[\rho]$ for some integer $n$. But this means that we have 
$$q_*(n\cdot [\rho])=n\cdot (q_*[\rho])=n\cdot [q\circ\rho]=n\cdot[\ast]=0$$
and can conclude that $q_*:\pi_4\mathbb{R}P^4\rightarrow\pi_4 S^4$ is zero.
Note at this stage that we can conclude the stronger statement that $q_*$ is the zero homomorphism in all degrees.
$(3)$ Assemble: By (1) any map $f:\mathbb{R}P^4\rightarrow S^3$ can be factored up to homotopy as 
$$f:\mathbb{R}P^4\xrightarrow{q} S^4\rightarrow S^3$$
where the second arrow may be inessential. It follows that $f_*:\pi_*\mathbb{R}P^4\rightarrow\pi_*S^3$ factors
$$f_*:\pi_*\mathbb{R}P^4\xrightarrow{q_*}\pi_*S^4\rightarrow\pi_*S^3$$
and we've seen in $(2)$ that the first homomorphism here is trivial.
