When $f\colon M\to N\times N$ satisfies that $f^{-1}(\Delta)$ is a ball? Let $M$, $N$ be smooth manifolds of dimension $m+n$ and $n$, respectively.
Suppose that $f\colon M\to N\times N$ is a smooth map and $f$ is transversal to the diagonal $\Delta=\{(x,x)\in N\times N\mid x\in N\}$. So, $f^{-1}(\Delta)$ will be sub-manifold of $M$.
Suppose that $f^{-1}(\Delta)$ is diffeomorphic to the $m$-dimensional Euclidean space. 
Now, my question is:
In this situation, are there any conditions on $M$ and $N$ that they should satisfy?
 A: Another way to look at the question is: when an $\mathbb R^m$ embedded if $M$ can be represented as $f^{-1}(\Delta)$ for such a transversal $f$? As commented before, the embedding must be closed (hence $M$ is not compact). Let us first look at the case $N=\mathbb R^n$. 
Prop.1 A submanifold $X\subset M$ diffeomorphic to $\mathbb R^m$ that is of the form $ \ f^{-1}(\Delta)\ $ for some $f:M\to\mathbb R^n\times\mathbb R^n$ must be a complete intersection in $M$.
Proof: Take $\delta:\mathbb R^n\times\mathbb R^n\to\mathbb R^n:(x,y)\mapsto x-y$ and define $F=(F_1,\dots,F_n)=\delta\circ f:M\to\mathbb R^n$. This $F$ is transversal to $0$ and the $F_i$'s are the regular equations making $X$ a complete intersection.$\ \square$ 
Conversely:
Prop.2  Let $X\subset M$ be a complete intersection of codimension $n$ diffeomorphic to $\mathbb R^m$. Then for every manifold $N$ of dimension $n$ there is $\ f:M\to N\times N$ transversal to $\Delta$ with $X=f^{-1}(\Delta)$.
Proof: Let $X$ be a complete intersection via $F:M\to\mathbb R^n$. Define $g:M\to\mathbb R^n\times\mathbb R^n$ by $g(x)=(0,F(x))$. Then $g$ is transversal to the diagonal $D\subset\mathbb R^n\times\mathbb R^n$ and $g^{-1}(D)=F^{-1}(0)=X$. Now pick any open embedding $h:\mathbb R^n\to N$, which exists because $n=\dim (N)$, and define $f=h\circ g$. This $f$ solves the problem.$\ \square$
We so see how complete intersections play here. Now this is a deep and difficult topic in differential topology. One first property of complete intersections is that their normal bundle is trivial, but there are submanifolds with trivial normal bundle which are not complete intersections. Akbulut-King found an exotic 16-sphere in $\mathbb R^{30}$ with trivial normal bundle that is not a complete intersection. In our case the submanifold is an $\mathbb R^n$, and its normal bundle is trivial because $\mathbb R^n$ is simply connected. We can get something from that:
Prop.3 Let $X\subset M$ be a closed submanifold of codimension $n$ diffeomorphic to $\mathbb R^n$. Then there exists a map $f:M\to\mathbb S^n\times \mathbb S^n$ transversal to $\Delta$ with $X=f^{-1}(\Delta)$.
Proof: Since the normal bundle of $X$ in $M$ is trivial, we find a tubular open neighborhood $U$: we have a diffeomorphism $\alpha:U\to\mathbb R^m\times\mathbb R^n$. Let $B$ and $\overline B$ denote the open and closed balls of radius $1$ centered at the origin. Since $X$ is closed in $M$ we can arrange things so that $g^{-1}(\mathbb R^m\times\overline B)$ is closed in $M$. Hence $V=g^{-1}(\mathbb R^m\times B)$ is another open neighborhood of $X$, with $\overline V=g^{-1}(\mathbb R^m\times\overline B)\subset U$. Now we consider two mappings:
(i) $\varphi:\mathbb R^n\to\mathbb S^n$, the inverse of the stereographic projection from the north pole $a$, which maps the origin to the south pole $-a$. Note that it maps the open ball $B$ to the lower open hemisphere.
(ii) $g:\mathbb S^n\to\mathbb S^n$, that collapses the upper closed hemisphere $H$ onto the north pole: $g(H_+)\equiv a$ and restricts to a diffeomorphism $\mathbb S^n\setminus H\to\mathbb S^n\setminus\{a\}$. 
With these at hand, let $\pi:\mathbb R^m\times\mathbb R^n\to\mathbb R^n\times\mathbb R^n:(x,y\mapsto (0,y)$ and set:
$$
f=(g\times g)\circ(\varphi\times\varphi)\circ\pi\circ\alpha:U\to\mathbb S^n\times\mathbb S^n
$$
One checks that this $f$ is transversal to $\Delta\subset\mathbb S^n\times\mathbb S^n$ and $f^{-1}(\Delta)=X$. Thus it remains to extend $f$ to $M$. But the construction gives in fact that $f^{-1}(\Delta)=f^{-1}(-a,-a)$ and $f\equiv (a,a)$ on $U\setminus\overline V$. Consequently we can extend $f$ be the constant value $(a,a)$, and this keeps all required properties. $\square$
This is far from a full answer, but I hope it gives some insight on the nature of the problem.
