Inspired by MSE/95760 I'm wondering whether the following is true: Let $\varphi : V \to W$ be an injective morphism between (affine) varieties. Does it follow that $\dim{V} \leq \dim{W}$?
I am not able to construct any counterexamples and I cannot see a way to translate chains of irreducible varieties in $V$ to $W$.
On
Someone posted an answer here but unfortunately deleted it because it was flawed. That is a pity because I believe that every mistake is educational and there were also several helpful comments. (Should it really be this easy to just delete answers on MSE?) That's why I am writing this answer. I want to retrieve everything that seems of importance for people in the future and let the previous author stay anonymous.
The claim was that the problem can be solved in a purely topological setting, i.e. let $f:X \to Y$ be an injective continuous map between topological spaces, then $\dim{X} \leq \dim{Y}$ for their Krull dimensions. We then try to lift a chain from $X$ to $Y$, that is let $$X_0 \subsetneq X_1 \subsetneq \dots \subsetneq X_n = X$$ be a chain of irreducible closed sets and look at $$\overline{f(X_0)} \subseteq \overline{f(X_1)} \subseteq \dots \subseteq \overline{f(X_n)}.$$ One can check that this is again a chain of irreducible closed sets and that's where the previous answer finished his proof.
There is a problem. It is not entirely clear that the inclusions in this new chain remain proper. In fact, it is wrong.
Let $X$ be any space of positive Krull dimension. Take any set $Y$ such that there exists an injective function $X \to Y$ and give $Y$ the indiscrete topology. So $\dim{Y} = 0$ and the map is continuous. That is a counterexample.
This example shows that the problem cannot be solved in a purely topological setting and requires some algebraic geometry.
The natural follow-up question is whether the dual problem (again, see MSE/95670) can be solved in a topological setting, i.e. given a surjective continuous map $f:X \to Y$ between topological spaces, do we get $\dim{X} \geq \dim{Y}$?
The answer is no, our previous idea can be applied again. Take a surjective map from a discrete topological space.
Let $f\colon X\to Y$ be an injective continuous map of sober topological spaces. Then $\dim(X)\leq\dim(Y)$, for it maps a sequence of non-trivial specialisations $x_n\leadsto x_{n-1}\leadsto\dots\leadsto x_0$ in $ X$, i.e., a sequence of distinct points such that $x_{i-1}\in\overline{\{x_i\}}$ for all $i=1,\dots,n$, to a sequence of non-trivial specialisations $f(x_n)\leadsto f(x_{n-1})\leadsto\dots\leadsto f(x_0)$ in $Y$.
In a bit more detail and without that terminology, if $X_0\subsetneq X_1\subsetneq \dots\subsetneq X_n$ is a chain of proper inclusions of irreducible closed sub-sets, then we get a corresponding sequence of generic points $x_i\in X_i$—i.e., $X_i=\overline{\{x_i\}}$ for all $i=0,\dots,n$—and so $x_{i-1}\in X_{i-1}\subsetneq X_i = \overline{\{x_i\}}$ for all $i=1,\dots,n$. Clearly, the $x_i$ are pair-wise distinct; in fact, $x_i\not\in X_{i-1}$. Consider $y_i:=f(x_i)$. Since $f$ is injective, those points are pair-wise distinct as well and since $Y$ is sober, so are their corresponding components $Y_i:=\overline{\{y_i\}}\subset Y$. Therefore, $Y_0\subsetneq Y_1\subsetneq \dots\subsetneq Y_n$. This is what's been missing to complete the proof attempted in the other answer.
Towards the follow up question in the OPs answer, cf. Stacks Project Tag 02JF.