Let $D(M)$ be the boundary of $M \times [0,1]$ (by smoothing corners, this can be understood as smooth). Then $f$ induces a smooth map $$ D(f): D(M) \to D(N)\, . $$ Further, $D(f)$ is a proper submersion of boundary-less manifolds so it's a smooth fiber bundle. Now pull this back along the inclusion $N \times 0 \subset D(N)$ to conclude that $f$ is a smooth fiber bundle.

Let $D(M)$ be the boundary of $M \times [0,1]$ (by smoothing corners, this can be understood as smooth). Then $f: M \to N$ induces a smooth map $$ D(f): D(M) \to D(N)\, . $$ Further, $D(f)$ is a proper submersion of boundary-less manifolds so it's a smooth fiber bundle. Now pull this back along the inclusion $N \times 0 \subset D(N)$ to conclude that $f$ is a smooth fiber bundle.

Let $D(M)$ be the boundary of $M \times [0,1]$ (by smoothing corners, this can be understood as smooth). Then $f$ induces a smooth map $$ D(f): D(M) \to D(N)\, . $$ Further, $D(f)$ is a proper submersion of boundary-less manifolds so it's a smooth fiber bundle. Now pull this back along the inclusion $M \times 0 \subset D(M)$ to conclude that $f$ is a smooth fiber bundle.

Let $D(M)$ be the boundary of $M \times [0,1]$ (by smoothing corners, this can be understood as smooth). Then $f$ induces a smooth map $$ D(f): D(M) \to D(N)\, . $$ Further, $D(f)$ is a proper submersion of boundary-less manifolds so it's a smooth fiber bundle. Now pull this back along the inclusion $N \times 0 \subset D(N)$ to conclude that $f$ is a smooth fiber bundle.