Theorem 1,1, Hirsch Differential topology Just getting started with Differential topology. Here is the first theorem I struggle with

Theorem 1.1. The set $\text{Imm}^r(M,N)$ of $C^r$ immersions is open in $C_S^r(M,N)$, $r \geq 1$.


Proof Since
$$
\text{Imm}^r(M,N) = \text{Imm}^1(M,N) \cap C^r(M,N)
$$
it suffices to prove this for $r = 1$. If $f : M \to N$ is a $C^1$ immersion one can choose a neighborhood $\mathcal{N}(f;\Phi,\Psi,K,\epsilon)$ as follows. Let $\Psi^0 = \left\{\psi_\beta,V_\beta\right\}_{\beta \in B}$ be any atlas for $N$. Pick an atlas $\Phi = \left\{\varphi_i,U_i \right\}_{i \in \Lambda}$ for $M$ so that each $U_i$ has compact closure, and for each $i \in \Lambda$ there exsts $\beta(i) \in B$ such that $f(U_i) \subset V_{\beta(i)}$. Put $V_{\beta(i)} = V_i$, $\psi_{\beta(i)} = \psi_i$, and $\Psi = \left\{\psi_i, V_i \right\}$. Let $K = \left\{ K_i \right\}_{i \in \Lambda}$ be a compact cover of $M$ with $K_i \subset U_i$.
The set
$$
A_i = \left\{ D(\psi_i f \varphi^{-1}_i)(x) : x \in \varphi(K_i) \right\}
$$
is a compact set of injective linear maps from $\mathbb{R}^m \to \mathbb{R}^n$. Since the set of all injective linear maps is open in the vector space $L(\mathbb{R}^m,\mathbb{R}^n)$ of all linear maps $\mathbb{R}^m \to \mathbb{R}^n$, there exists $\epsilon_i > 0$ such that $T \in L(\mathbb{R}^m,\mathbb{R}^n)$ is injective if $\left\|T - S \right\| < \epsilon_i$ and $S \in A_i$. Set $\epsilon = \left\{ \epsilon_i \right\}$. It follows that every element of $\mathcal{N}^1(f ; \Phi, \Psi, K, \epsilon)$ is an immersion.


*

*I'm not really sure what the author is trying to prove. I think he wants to show that $\text{Imm}^1(M,N)$ has sets of the $\mathcal{N}^1(f ; \Phi,\Psi,K,\epsilon)$ as a subbasis. Is this correct?


*The picking of atlas $\Psi^0$ is possible by definition of manifold. However I'm not sure why he's able to pick an atlas $\Phi = \left\{\varphi_i, U_i \right\}$ with the mentioned features, namely  such that the closure of $U_i$ is compact and such that $f(U_i) \subset V_{\beta(i)}$. For the latter I think since $f$ is an immersion, in particular is homeomorphic and hence continuous. For fixed $B$ each $V_{\beta(i)}$ there's an open $U_i$ in $M$ such that $f(U_i) \subset V_{\beta(i)}$. For the compactness I think we would need to assume the manifold $M$ is compact, but I don't think the author explicitly assumes that so I'm not sure.


*Not sure why $A_i$ is compact.
 A: Before we get started, let me stress that $M$ and $N$ are finite-dimensional (paracompact Hausdorff) manifolds, but they may very well be non-compact.
Let's go through your questions one by one.
Ad 1. In the end, we want to show that $\mathrm{Imm}^r(M,N)$ is open in $\mathrm{C}^r(M,N)$ where the latter carries the strong topology.
By the first remark in the proof, it is actually enough to show that $\mathrm{Imm}^1(M,N)\subset \mathrm{C}^1(M,N)$ is open.
Recall that the strong topology on $\mathrm{C}^1(M,N)$ is defined by the basis of opens $\mathcal N^1(f;\Phi,\psi,K,\epsilon)$.
Thus, to show that $\mathrm{Imm}^1(M,N)\subset \mathrm{C}^1(M,N)$ is open, it is sufficient to show that for an arbitrary $f$ and a suitable choice of $\Phi,\psi,K$ and $\epsilon$, the open neighbourhood $\mathcal N^1(f;\Phi,\Psi,K,\epsilon)$ of $f$ (in $\mathrm{C}^1(M,N)$) is contained in $\mathrm{Imm}^1(M,N)$, i.e., consists of immersions only. This is what the author is proving.
Ad 2. We start with any atlas $\Psi^0=\{\psi_\beta,V_\beta\}_{\beta\in B}$ on $N$. The claim is that there exists an atlas $\Phi=\{\varphi_i,U_i\}_{i\in\Lambda}$ on $M$ with the following properties:

*

*The covering $\{U_i\}_{i\in\Lambda}$ is locally finite.

*For every $i\in \Lambda$, there exists a $\beta(i)\in B$ such that $f(U_i)\subset V_{\beta(i)}$.

*Every $U_i$ is relatively compact (i.e., $\overline{U_i}$ is compact).

*There is a covering $M=\bigcup_{i\in\Lambda}K_i$ by compact sets $K_i$ contained in $U_i$.

For every point $x\in M$, we pick a $b(x)\in B$ such that $f(x)\in  V_{b(x)}$. Since $f$ is continuous (at $x$), there exists an open neighbourhood $W_x$ such that $f(W_x)\subset V_{b(x)}$. Shrinking the $W_x$, if necessary, we may suppose that they form an atlas $\{\omega_x,W_x\}$. Shrinking $W_x$ further, if necessary, we may assume that all $W_x$ are relatively compact; this is possible since $M$ is locally compact and Hausdorff. Note that shrinking doesn't change the fact that $f(W_x)\subset V_{b(x)}$. The covering $\{W_x\}_{x\in M}$ is not locally finite, so we have to take a locally finite refinement, which is possible since $M$ is paracompact. That is, we get an open covering $\{U_i\}_{i\in \Lambda}$ and a map $\rho\colon\Lambda\to M$ such that $U_i\subset W_{\rho(i)}$. Letting $\varphi_i$ denote the restriction of $\omega_{\rho(i)}$ to $U_i$, we get an atlas $\{\varphi_i,U_i\}_{i\in\Lambda}$ which satisfies 1. by construction, 3. since $U_i\subset W_{\rho(i)}$ and 2. through $\beta:=b\circ \rho$ since $f(U_i)\subset f(W_{\rho(i)})\subset V_{b(\rho(i))}$. Finally, we may apply the Shrinking Lemma to obtain a refinement $\{U'_i\}_{i\in\Lambda}$ such that $U'_i\subset \overline{U'_i}\subset U_i$ for all $i\in\Lambda$. Denoting $K_i:= \overline{U'_i}$ (and forgetting about the $U'_i$), we get a compact cover $K=\{K_i\}_{i\in\Lambda}$ as desired in 4.
Ad 3. Since $f$ is $\mathrm{C}^1$ and we are dealing with $\mathrm{C}^1$-atlasses, $\psi_i f\varphi^{-1}$ is $\mathrm{C}^1$ as well (on $\varphi(U_i)$ anyways). Therefore, the map $D(\psi_i f\varphi_i^{-1})\colon \varphi(U_i)\to L(\mathbb R^m,\mathbb R^n)$, $x\mapsto D(\psi_i f\varphi_i^{-1})(x)$, is continuous. Since $K_i$ is compact, so is its continuous image $\varphi_i(K_i)$; thus, $A_i$ is compact as well, being the continuous image of $\varphi(K_i)$ unter $D(\psi_i f\varphi_i^{-1})$.
