Dualizing from co-invariants to invariants I'm reading a paper where at some point the author is trying to gain some information about the dimension of some invariant space of some homology group of a space. For this, they prove the corresponding claim with respect to the co-invariants of the cohomology of said space. Then they just say "by dualizing, we obtain [the information about invariants of homology]". 
I think I've summarized the way they obtain this by the following claim (which may have too many hypotheses, I'm just trying to have as many as I can to make sure I'm not missing something):

Let $G$ be a finite elementary abelian $p$-group, $M$ a finitely generated $\mathbb{F}_p[G]$-module. Then the restriction map $\hom(M,\mathbb{F}_p)\to \hom(M^G,\mathbb{F}_p)$ factors through an isomorphism $\hom(M,\mathbb{F}_p)_G\to  \hom(M^G,\mathbb{F}_p)$; i.e. the kernel of the restriction map is $I(G)\hom(M,\mathbb{F}_p)$

With $\hom$ meaning simply $\mathbb{F}_p$-maps, given the usual $G$-action ($\mathbb{F}_p$ has the trivial $G$-action) and $I(G)$ is the augmentation ideal. 
However I don't know if that claim is true, and it actually seems not to be true. To be fair, I only need the restriction map to be injective, but it is automatically surjective since we're over vector spaces. 
Let $f$ have zero restriction on $M^G$, then what I could find was that $P\cdot f = 0$, where $P=\displaystyle\sum_g g$ is the usual projection, but it's not really helping me (I can't divide by the order of $G$, so I can't have $f= (1-\frac{1}{|G|}P)f$ which would work fine)
I don't have tons of examples at hand, it works well for trivial modules and for free modules
So my first question is :

Is the above claim true ? If not, what would be a counterexample ? If yes, could you give a hint to prove it ?

If the answer is yes, you may stop reading, because the rest of the question is about the claim being false. 
Now if the above claim is false in general, it means I don't understand what's happening in the paper and I now need to add more information about it. The claim I'm referring to is the last part of the proof of Thm 1.1 in The free rank of symmetry of $(S^n)^k$, by Adem and Browder. 
They're dealing with a space $X$ with a bunch of hypotheses and a free action of $G$ a finite $p$-elementary abelian group on this space, $H$ is the subgroup of elements that act $\mathbb{Z}_{(p)}$-homologically trivially, and they proved $\mathrm{rk}(H)\leq \dim H^n(X;\mathbb{F}_p)_G$. I'm fine with the proof until that point; but then they state "by dualizing, we get $\mathrm{rk}(H)\leq \dim H_n(X;\mathbb{F}_p)^G$". 
The claim I extracted from that comes from the universal coefficient theorem and my understanding of the phrase "by dualizing". Indeed we have (over a field) $H^n(X;\mathbb{F}_p) = \hom (H_n(X;\mathbb{F}_p), \mathbb{F}_p)$ and this isomorphism being natural, it commutes with the action of $G$, in other words it is an isomorphism of $G$-modules. So what I understand by dualizing is relating $H^n(X;\mathbb{F}_p)_G$ and $H_n(X;\mathbb{F}_p)^G$ by means of this isomorphism, that is, relate $\hom(M,\mathbb{F}_p)_G$ and $\hom(M^G,\mathbb{F}_p)$ for $M=H_n(X;\mathbb{F}_p)$ : this is where the above claim comes from.
So in case the claim is not true, my second question is :

What is meant here with "by dualizing", and how does it help us get the inequality ?

 A: EDIT : There is, in fact, a much simpler argument that I found while discussing with friends. This is probably what was meant by the authors. I'll wrire it down here at the beginning (but I will leave the complicated argument, as I think it can still be interesting)
The easy argument is as follows : note that $\hom(N_G, \mathbb F_p) \cong \hom(N, \mathbb F_p)^G$ naturally, immediately by definition.
Apply this to $N=\hom(M, \mathbb F_p)$ to get $\hom(M^*_G, \mathbb F_p) \cong \hom(M^*, \mathbb F_p)^G = (M^{**})^G \cong M^G \cong \hom(\hom(M^G, \mathbb F_p), \mathbb F_p)$.
Therefore by applying $\hom(-, \mathbb F_p)$, and duality in finite dimensioms, we get the desired isomorphism (one may check that it's the correct ome, but it's enough that there is one to apply a dimension argument in what follows).
Now, the old, complicated answer :
For projective (hence free, given the hypotheses) $M$, the map in question is an isomorphism, because it is surjective and we can just count dimensions over $\mathbb F_p$ to see that they're the same (for projective $M$, $\dim M^G = \frac{\dim M}{|G|}$).
Now I would like to go from the fact that this holds for f.g. projective modules to general f.g. modules, by using a homological argument as follows: let $M$ be our f.g. module, and $P\to M$ a surjective map from a f.g. projective $P$. 
Let $N = \ker(P\to M)$. 
$\hom$ will denote morphisms of vector spaces over $\mathbb F_p$, not $G$-modules. So we have a short exact sequence of $G$-modules $0\to \hom(M, \mathbb F_p)\to \hom(P,\mathbb F_p)\to \hom(N,\mathbb F_p)\to 0$, and if we take the coinvariants, given that $P$ and therefore $\hom(P,\mathbb F_p)$ are projective, we have an exact sequence  $$0\to H_1(G, \hom(N,\mathbb F_p)) \to \hom(M,\mathbb F_p)_G \to \hom(P, \mathbb F_p)_G\to \hom(N,\mathbb F_p)_G\to 0$$
On the other hand, if we take invariants, we have an exact sequence $0\to N^G\to P^G\to M^G\to H^1(G,N)\to 0$ ($P$ is projective, hence injective, because we're over $\mathbb F_p[G]$), and now we may take $\hom(-,\mathbb F_p)$ which is exact to get the following exact sequence : $$0\to \hom(H^1(G,N), \mathbb F_p)\to \hom(M^G,\mathbb F_p) \to \hom(P^G, \mathbb F_p)\to \hom(N^G,\mathbb F_p)\to 0$$
Of course the comparison map I defined in my question was natural (it's simply a restriction) so that we can compare these two exact sequences to get a big commutative diagram : 
$$\require{AMScd} \begin{CD}0 @>>> H_1(G, \hom(N,\mathbb F_p)) @>>> \hom(M,\mathbb F_p)_G @>>>  \hom(P, \mathbb F_p)_G@>>> \hom(N,\mathbb F_p)_G@>>>  0 \\
@VVV @VVV @VVV @VVV @VVV @VVV \\
0 @>>> \hom(H^1(G,N), \mathbb F_p) @>>>\hom(M^G,\mathbb F_p) @>>> \hom(P^G, \mathbb F_p)@>>> \hom(N^G,\mathbb F_p)@>>> 0\end{CD}$$
where the map on the level of (co)homology is obtained simply from the universal property of kernels. 
I claim that the leftmost map is surjective. Indeed, the map $\hom(M,\mathbb F_p)_G\to \hom(M^G,\mathbb F_p)$ is known to be surjective, so if we take $x\in \hom(H^1(G,N), \mathbb F_p)$, push it to $\hom(M^G,\mathbb F_p)$, lift it to $\hom(M,\mathbb F_p)_G$, then push it to $\hom(P,\mathbb F_p)_G$, then down to $\hom(P^G,\mathbb F_p)$, we get that the lift we chose was actually in $H_1(G, \hom(N,\mathbb F_p))$, which implies that $x$ is in the image, i.e. the map is surjective. 
Now one can check that the kernel of our desired map is precisely the kernel of the leftmost map (this is a simple diagram chase). Let's check that the two terms of this map have the same dimension. 
Over $\mathbb F_p[G]$, projective is equivalent to injective, so I may take a finitely generated injective resolution $0\to N\to P_0\to P_1\to P_2$, which yields a projective resolution $P_2^* \to P_1^* \to P_0^*\to N^*$ (where $^*$ just means dual over $\mathbb F_p$), of which I may take the coinvariants $(P_2^*)_G\to (P_1^*)_G\to (P_0^*)_G$. The cohomology of this complex is $H_1(G,\hom(N,\mathbb F_p))$. 
On the other hand, given this injective resolution, if I take the invariants, I get $P_0^G\to P_1^G\to P_2^G$. Taking $\hom(-,\mathbb F_p)$ we get $(P_0^G)^*\to (P_1^G)^* \to (P_2^G)^*$. The cohomology of this (which of course commutes with $(-)^*$, as that functor is exact) is precisely $H^1(G,N)^*$. 
But now, given the earlier observation that our natural morphism was an iso on projectives, we see that our two complexes are isomorphic ! It follows that they have the same cohomology - $H^1(G,N)^* \cong H_1(G,N^*)$. 
Given that they have the same dimension, surjectivity of the leftmost map is enough to imply injectivity. 
All in all, the natural map was injective, hence an isomorphism. 
Pfew. A few remarks : 
1- It's possible that I'm missing an obvious argument that would show $N\cong N^*$ as $G$-modules, in which case the last argument could be bypassed by the Künneth formula, to make everything easier. But I'm not sure that this is the case, and don't see any argument for it (in fact I think it's not true in general but don't have any counterexample) 
2- I wrote this argument at the same time as I was working it out, so it's not polished at all, and it's very likely that it can be made simpler (or that it contains mistakes - I hope not, even though I wrote it late, I spotted a few mistakes along the way and so I think I still have some ability to discern things)
3- There may very well be a way more elementary argument that consists in taking $f$ in the kernel and by a trick finding that it's in $I(G)\hom(M,\mathbb F_p)$. I don't see it at the moment. 
4- In the last argument, it's very likely that the leftmost map is precisely the one I exhibit when I prove that the two cohomology groups are isomorphic. It's not necessary for the argument to have that they're the same map, and frankly, given how tired I am, I don't want to check whether they are. 
