If $f\colon S^n \longrightarrow S^m$ continuous, $m\geq n$. I am trying to show that if $f\colon S^n\longrightarrow S^m$ is a continuous map and $f(-x)=-f(x)$, then $m\geq n$.
First of all, the cohomology ring $H^{*}(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z}$) is $$\frac{\mathbb{Z}_{2}[x]}{(x^{n+1})},$$ and using the Gysin sequence we obtain that $H^{i}(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})=\langle e(E)^{i}\rangle$, where $e(E)$ is the Euler class of the canonical $1$-dimensional vector bundle of $\mathbb{R}P^n$.
Secondly, since $f(-x)=-f(x)$, $f$ descends to a well-defined map $f\colon \mathbb{R}P^n\longrightarrow \mathbb{R}P^m$.
Then, $f^{*}\colon H^{1}(\mathbb{R}P^m;\mathbb{Z}/2\mathbb{Z})\longrightarrow H^{1}(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})$. Since $$H^{1}(\mathbb{R}P^i;\mathbb{Z}/2\mathbb{Z})=\mathbb{Z}/2Z,$$ there are two options for $f^{*}$: it is $0$ or it is bijective.
If I prove that it cannot be $0$, I would be done. However, I do not know how to show that $f^{*}$ can't be $0$. Can anyone help me, please?
$\mathbf{EDIT:}$
$f^{*}(e(E))=e(f^{*}(E))$ where $f^{*}(E)$ is the pullback. That is, $$f^{*}(E)=\{([y],([x],v)\mid [y]\in \mathbb{R}P^n, x\in \mathbb{R}P^m, [f(y)]=[x], v=\lambda x \text{ for some } \lambda \in \mathbb{R}\}.$$
I think that I have proved that $f^{*}(E)$ is isomorphic to the $1$-dimensional vector bundle of $\mathbb{R}P^n$ by sending $([y],([x],\mu f(y)))$ in $f^{*}(E)$ to $([y],\mu y)$ in the canonical vector bundle. That map is bijective and there is a linear isomorphism between the fibres. However, I have not used that $f(-x)=-f(x)$, so I believe that something has to be wrong...
 A: You can prove that $f^*$ must be an isomorphism by using that $f_{\#}$ must be an isomorphism on fundamental groups and then using both Hurewicz and the Universal Coefficients Theorem.
To prove that $f_{\#}$ is an isomorphism on fundamental groups, you indeed need that $f(x)=-f(-x)$. You pick a point $p$ and a path $\gamma$ taking $p$ to $-p$ on the sphere $S^n$. The path $f \circ \gamma$ does the same thing (takes a point $p'$ to $-p'$) by the property that $f(x)=-f(-x)$, and thus the map induced in projective spaces induces an isomorphism in fundamental groups (it is taking a non-trivial loop to a non-trivial loop, and the fundamental group is $\mathbb{Z}_2$).

However, there is a more elementary approach (which can be seen in Bredon):
Given a two-sheeted covering $p:X \to Y$ (in this case, $p_n: S^n \to \mathbb{R}P^n$), there is a long exact sequence
$$\cdots \to H_p(Y;\mathbb{Z}_2) \stackrel{t_*}{\to} H_p(X,\mathbb{Z}_2) \stackrel{p_*}\to H_p(Y;\mathbb{Z}_2) \stackrel{\partial_*}\to H_{p-1}(Y;\mathbb{Z}_2) \to \cdots,  $$
which is obtained from the short exact sequence of complexes
$$0 \to \Delta_p(Y;\mathbb{Z}_2) \stackrel{t}{\to} \Delta_p(X,\mathbb{Z}_2) \stackrel{p_{\Delta}}\to \Delta_p(Y;\mathbb{Z}_2) \to 0 ,$$
where the only map which needs explaining is $t$: this is the map taking $\sigma$ to $\sigma +a \circ \sigma,$ where $a$ is the only deck transformation which is not the identity. You can check that this sequence is indeed exact.
For your case of $S^n$ and $\mathbb{R}P^n$, we have
$$0 \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \to  H_m(S^m,\mathbb{Z}_2) \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \to H_{m-1}(\mathbb{R}P^m;\mathbb{Z}_2) \to \cdots,  $$
$$\cdots \to H_1(\mathbb{R}P^m;\mathbb{Z}_2) \to H_0(\mathbb{R}P^m;\mathbb{Z}_2)  \to   H_0(S^m,\mathbb{Z}_2) \to H_0(\mathbb{R}P^m;\mathbb{Z}_2) \to 0.  $$
We have that the above sequence is
$$0 \to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}{\to}  H_m(S^m,\mathbb{Z}_2) \stackrel{0}\to H_m(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}\to H_{m-1}(\mathbb{R}P^m;\mathbb{Z}_2) \to \cdots,  $$
$$\cdots \stackrel{0}\to H_1(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{\simeq}{\to} H_0(\mathbb{R}P^m;\mathbb{Z}_2) \stackrel{0}\to  H_0(S^m,\mathbb{Z}_2) \stackrel{\simeq}\to H_0(\mathbb{R}P^m;\mathbb{Z}_2) \to 0.  $$
This is trivial if you know the homology of $\mathbb{R}P^n$ and $S^n$ (the first map is injective, but since it is a map from $\mathbb{Z}_2$ to $\mathbb{Z}_2$ it must be an isomorphism etc), but could also be proven without resorting to knowing the explicit homology of $\mathbb{R}P^n$, if you go down to using the definitions of the maps.
This tells us that each $\partial_*$ is an isomorphism. 
Let's now assume that $n>m$. Naturality of $\partial_*$ tells us that the following diagram commutes for all $i$
$$\begin{array}{ccccccccc} H_i(\mathbb{R}P^n;\mathbb{Z}_2) & \xrightarrow{\partial_*} & H_{i-1}(\mathbb{R}P^n;\mathbb{Z}_2)  \\
\downarrow{f_*} & & \downarrow{f_*} \\
H_{i}(\mathbb{R}P^m;\mathbb{Z}_2) & \xrightarrow{\partial_*} & H_{i-1}(\mathbb{R}P^m;\mathbb{Z}_2).\end{array}$$
By induction and since $\partial_*$ is an isomorphism up to $m$ (both above and below, since $n>m$) by what we've seen above and also since $f_*$ is obviously a isomorphism when $i-1=0$, it follows that $f_*$ is an isomorphism up to when $i=m$. 
But this contradicts the following commutative diagram
$$\begin{array}{ccccccccc} H_m(\mathbb{R}P^n;\mathbb{Z}_2) & \xrightarrow{t_*} & H_{m}(S^n;\mathbb{Z}_2)  \\
\downarrow{f_*} & & \downarrow{f_*} \\
H_{m}(\mathbb{R}P^m;\mathbb{Z}_2) & \xrightarrow{t_*} & H_{m}(S^m;\mathbb{Z}_2),\end{array}$$
since going from top left to down right is a composition of isomorphisms of non-trivial groups, whereas the top right is $0$.
A: The condition implies that the map descends to a map $f\colon PR^n\to RP^m$ that induces isomorphism on the fundamental groups.  Composition with inclusion in $K(\pi,1)=RP^\infty$ must induce isomorphism in the homology ring up to dimension $n$.  Hence $m\geq n$.
Without using the terminology of Eilenberg-MacLane spaces $K(\pi,1)$, one can argue by contradiction as follows.  Suppose $m<n$.  Extend the map $f$ to a map $f\colon RP^n \to RP^n$ by using a standard inclusion $RP^m \hookrightarrow RP^n$.  Show that this map induces isomorphism in $H_k(RP^n,Z_2)$ for all $k\leq n$. This gives the desired contradiction since the map factors through $RP^m$.
