The Arnol'd conjecture for symplectomorphisms sufficiently close to the identity In studying symplectic geometry, a relatively easy corollary of Weinstein's Lagrangian Neighbourhood theorem is the following.

On a closed symplectic manifold $(M,\omega)$ with $H^1_{\text{dR}}(M)=0$, every symplectomorphism sufficiently $C^1$-close to the identity has at least $\text{Crit}(M)$ fixed points.

In the proof of this, it is claimed that if $\phi$ is sufficiently $C^1$ close to the identity, the graph of $\phi$ is contained in a given open neighbourhood $\mathcal{U}$ of the diagonal $\{(x,x)| x\in M\}$. Why is this claim actually true?
For filling in the details, I would like to find a $C^1$ neighbourhood $\mathcal{N}$ of the identity such that if $\phi\in\mathcal{N}$, then $\text{Graph}(\phi)\subset \mathcal{U}$. Intuitively, it is very clear that for $\phi(x)$ sufficiently close to $x$, we have that $\text{Graph}(\phi)$ is contained in $\mathcal{U}$. However, how can I choose such an open neighbourhood $\mathcal{N}$ such that it actually works that the graph of each $\phi$ in this open neighbourhood is contained in $\mathcal{U}$? I know from Page 37 in Hirsch, Differential topology that the group of $C^1$ diffeomoprhisms $M\to M$ is open in $C^1(M)$, but I could not find the result in there any further.
 A: Note that if $g \in C^1(L;T^*L)$ is an embedding such that $\pi \circ g$ is an embedding, then the map $C^1(L;T^*L)\to C^1(L;L)$ given by $h \mapsto \pi \circ h$ takes embeddings near $g$ to embeddings. Indeed, since embeddings are open, we can take a neighbourhood $\mathcal{U}_1$ of $g$ such that all of its elements are embeddings. Then, we intersect this with a neighbourhood $\mathcal{U}_2$ for which the elements don't have derivative sending into the kernel of $d\pi$. (Since the set 
$$\{h : \exists x \in M, 0 \neq v  \in T_xM \text{ with } d\pi(dh(v))=0\}$$ is closed in $C^1(L; T^*L),$ this is possible. If this isn't clear, you can pick $h_n \to h$ together with chosen $x_n$ and normalized $v_n$ after some choice of metric on $L$ and use compactness of $L$ and of the unit sphere.) The image of $\mathcal{U}=\mathcal{U}_1 \cap \mathcal{U}_2$ therefore consists of maps $h$ such that $\pi \circ h$ are embeddings, which must  in fact be diffeomorphisms by the inverse function theorem.
Any $h \in \mathcal{U}$ will be able to be "reparametrized" to be a section. Indeed, letting $\alpha : = (\pi \circ h)^{-1}$, we have that $h' = h \circ \alpha$ is such a reparametrization. Note that it is obviously a section, since $\pi \circ h' = \pi \circ h \circ (\pi \circ h)^{-1}=\mathrm{Id}$, which is the definition of being a section.
Now just apply all of the above to the case where $g$ is the embedding of $L$ as the zero section. Let $f:U \to U'$ be the symplectomorphism given by the Weinstein neighbourhood theorem, where $U$ is a neighbourhood of the graph $L$, which is a Lagrangian submanifold, and $U'$ is a neighbourhood of the zero section of $T^*L$. Then there is a map $f_{\#}:C^1(M;M\times M)^{U} \to C^1(L;T^*L)$ given by composition and extension. (The superscript means that we are restricting to those maps that take $M$ inside $U$.) This map is continuous, since $f$ is a diffeomorphism. So picking $\mathcal{V}$ to be $f_{\#}^{-1}(\mathcal{U})$ and considering $\iota^{-1}(\mathcal{V})$, where
\begin{align*}
\iota: C^1(M; M) &\to C^1(M;M \times M) \\
h &\mapsto \left(x \mapsto \left(x,h\left(x\right)\right)\right),
\end{align*} 
which is again continuous, you have the neighbourhood you want.

For the sake of completeness and future reference: in order to complete the argument to prove the Arnold conjecture in this case, note that graphs of symplectomorphisms are Lagrangian submanifolds of the product. So the image of the graph of a symplectomorphism which is in the neighbourhood constructed above will yield a Lagrangian submanifold which is the geometric realization of a section of the cotangent bundle, i.e. a one-form $\mu$.
This one-form $\mu$ is closed, since a geometric realization of a one-form is Lagrangian in the cotangent bundle if and only if the one-form is closed. (This is a well-known fact, but can be seen in page 17 here) By the hypothesis of $H^1_{dR}(M)=0$, it is exact, i.e. $\mu=dk$ for some real function $k: L \to \mathbb{R}$. The critical points of $k$ are then precisely the intersections of the geometrical realization of $\mu$ with the zero section, which in turn are precisely the points for which $f_{\#}(\iota(\phi)))(x)=f_{\#}(\iota(\mathrm{Id}))(y)$, i.e., $f((x,\phi(x)))=f((y,y))$ for some $x$ and $y$. Since $f$ is bijective, we must have $(x,\phi(x))=(y,y)$, and then $x=y$ and $\phi(x)=x$. It follows that the fixed points of $\phi$ are in bijection with the critical points of $k$, thus the fixed points of a $\phi$ which is in $\iota^{-1}(\mathcal{V})$ are bounded from below by the minimum number of critical points which a smooth function in $M$ can have. 
