properly embedded submanifold

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Definition: Let $Y$ a smooth manifold, a proper submanifold of $Y$ is a submanifold $(X,\phi)$ such that $\phi:X\to Y$ id s proper map.

I will appreciate some hint to solve the following exercise Question: Let $f:M\to N$ a submersion and let $P\subset N$ a proper submanifold, show that $f^{-1}(P)\subset M$ is a proper submanifold.

Here is what I tried

Because $P\subset N$ a proper submanifold, then the inclusion map $i_N:P\to N$ is proper.

Let us prove that the inclusion map $i_M:f^{-1}(P)\to M$ is proper. Let $K\subset M$ a compact set, we have to proof that $i_M^{-1}(K)=K\cap f^{-1}(P)$ is compact.

Edited

Let $(V_i)_{i\in I}$ an open cobering of $i_M^{-1}(K)=K\cap f^{-1}(P)$ on $ f^{-1}(P)$, then there are open sets $U_i$ of $M$ such that $$V_i=U_i\cap f^{-1}(P),$$ then $$K\cap f^{-1}(P)\subset\bigcup_{i\in I} V_i=\bigcup_{i\in I} \left( U_i\cap f^{-1}(P)\right)=\left(\bigcup_{i\in I}U_i \right)\cap f^{-1}(P).$$

Hence \begin{equation}\label{1081} K\cap f^{-1}(P)\subset\left(\bigcup_{i\in I}U_i \right)\cap f^{-1}(P).....(*) \end{equation}

On the other hand, \begin{equation}\label{1082} f^{-1}(f(K)\cap P)=f^{-1}(f(K))\cap f^{-1}(P) .....(**) \end{equation}

and $$K\cap f^{-1}(P)\subset f^{-1}(f(K))\cap f^{-1}(P),\qquad(\mbox{ since } K\subset f^{-1}(f(K))),$$ then in em $(**)$ we get

$$K\cap f^{-1}(P)\subset f^{-1}(f(K))\cap f^{-1}(P)= f^{-1}(f(K)\cap P)$$

Here is where I get stuck.

What I want to get is that $$f(K)\cap P\subset\displaystyle\bigcup_{i\in I}f(U_i)\cap P$$ and since $i_N:P\to N$ is proper, and $f(K)$ is compact, then $f(K)\cap P$ is compact on $P$, and I get a finite subcovering and finish the proof.

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I suppose that you are working with manifolds, and everything is separated. Let $(x_n)$ be a sequence of $f^{-1}(P)\cap K$ wich converges towards $x$, $x\in K$ since $K$ is compact. The sequence $(f(x_n))$ is a sequence of $f(K)\cap P$ since $P$ is proper and $f(K)$ compact we deduce that $f(K)\cap P$ is compact and $f(x)=lim_nf(x_n)\in f(K)\cap P$. This implies that $f(x)\in P$ and $x\in f^{-1}(P)$, we deduce that $x\in K\cap f^{-1}(P)$. You don't need submersion here.