# Image of a projective variety is a projective variety

I am using Karen E. Smith et al.'s Invitation to Algebraic Geometry, and was wondering the following: if $$\phi : V \subseteq \mathbb{P}^n \to W \subseteq \mathbb{P}^m$$ is a morphism of quasiprojective varieties (as they define it in the textbook, so without using any commutative algebra!), and $$V$$ is Zariski closed in $$\mathbb{P}^n$$, in other words, it is a projective variety, is $$\phi(V)$$ then a Zariski closed subset of $$W$$?

I have looked through this site and found a few questions that look very similar, like Image of morphism of projective varieties is projective variety and Is the image of a projective variety projective? . However, it appears that they contradict each other. Which is probably because they use different definitions of what projective means?

I was wondering if the statement is true in the context of the definitions handled by Karen E. Smith et al. and how to prove it.

What I have tried:

If $$V \subseteq \mathbb{P}^n$$ is Zariski closed, then I know it is given by the vanishing of finitely many homogeneous polynomials in the variables $$x_0, \ldots, x_n$$, say, $$p_1, \ldots, p_K$$.

The definition of morphism given in Smith et al. is very local: in each $$p \in V$$ we find an open subset $$U_p \subseteq V$$ such that there exist homogeneous polynomials $$F_0^{(p)}, \ldots, F_m^{(p)}$$ in the variables $$x_0, \ldots, x_n$$ such that $$\phi$$ agrees on $$U^{(p)}$$ with the map $$q \mapsto [F_0^{(p)}(q) : \cdots : F_m^{(p)}(q)].$$

Because $$V$$ is compact in the Zariski topology, we can pick finitely many $$U^{(p_1)}, \ldots, U^{(p_L)}$$ that cover $$V$$. I suppose I should try to find equations for $$\phi(V)$$ now. But I do not see an obvious way to combine the $$F_j^{(p_l)}(x_0, \ldots, x_n)$$'s and the $$p_k(x_0, \ldots, x_n)$$'s to find a set of homogeneous equations for $$\phi(V)$$.

There's no contradiction between the two linked questions: in the first link, the target of the morphism is a subvariety (or subscheme) of $$\Bbb P^n$$, while in the second link, the target is not assumed to live anywhere in particular. This matters to the end result - the underlying claim here which is true is the following:
Theorem. Let $$X\subset \Bbb P^n$$ be a closed subvariety. Then any map of varieties $$f:X\to Y$$ is closed.
If $$Y\subset\Bbb P^m$$, you can consider the composition $$X\to Y \to \Bbb P^m$$ and so the image of $$X$$ is closed in $$\Bbb P^m$$ and therefore a subvariety. If $$Y\not\subset\Bbb P^m$$, then you can't make this conclusion.