= Solution
Suppose the map were nonconstant. Since <projective space> is proper, its image $Z$ is a closed projective subvariety of $\mathbb P^n$, of positive dimension $r\le n$. Fix $y\in Z$. Regard the map as surjective onto $Z$; the <fiber dimension theorem> just proved gives
$$
\dim\phi^{-1}(y)\ge m-r\ge m-n>0.
$$
Choose a target hyperplane $H$ not containing $y$. It meets $Z$: the <affine cone> over $Z$ has dimension $r+1\ge2$, and the hyperplane equation vanishes at its vertex but not identically. By the <principal hypersurface dimension lemma> its zero set on the cone has dimension $r\ge1$, so it contains a nonzero cone point and hence a point of $Z\cap H$.
The inverse image $D=\phi^{-1}(H)$ is therefore nonempty and proper. Locally the hyperplane equation is a single regular equation, after trivializing the pulled-back <line bundle>. On the irreducible smooth source it is not identically zero, so the same hypersurface dimension lemma shows that each component of $D$ has dimension $m-1$. Choose one component $D_0$. Its cone is a codimension-one irreducible closed subset of affine $(m+1)$-space, whose prime ideal is principal because the <polynomial ring> is a <unique factorization domain>. Thus $D_0$ is a <projective hypersurface> defined by a homogeneous polynomial $P$ of positive degree.
Choose a positive-dimensional <irreducible component> $F$ of $\phi^{-1}(y)$. Since $y\notin H$, $F$ is disjoint from $D_0$. On the cone over $F$, the equation $P$ is consequently nonzero everywhere except at the vertex. This contradicts the hypersurface dimension lemma: the cone has dimension at least two, so its nonempty proper zero set for $P$ has dimension at least one and cannot consist just of the vertex. Therefore \b[$\boxed{\phi\text{ is constant}}$]. The argument proves <morphisms from projective space to smaller projective space are constant> by combining the fibre bound with the forced intersection of a positive-dimensional <projective variety> and a hypersurface.
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