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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 givesChoose a target hyperplane $H$ not containing $y$. It meets $Z$: the affine cone over $Z$ has dimension $r+1\ge 2$, 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 $r\ge 1$, so it contains a nonzero cone point and hence a point of $Z\cap H$.

The inverse image $D={\varphi }^{-1}(H)$ 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 ${\varphi }^{-1}(y)$. 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 $\boxed{\varphi \text{ is constant}}$. 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.