Given a flat morphism of schemes $f:X\rightarrow Y$ of finite type over a field $k$, I would like to understand how we can "reduce" to the case where $Y$ is irreducible. The context, if needed, is the proof of point (b) of Proposition III.10.1 of Hartshorne's book. There, he states that the conditions
if $X' \subseteq X$ and $Y' \subseteq Y$ are irreducible components such that $f(X')\subseteq Y'$, then $dim(X')=dim(Y')+n$
and
for any point $y\in Y$, every irreducible component of the fibre $X_y$ of $f$ has dimension $n$
are equivalent. To justify this, he refers to a theorem that gives the result in the case where $Y$ is irreducible, and gives no justification as for the general case.
So, to reduce to the irreducible case, I thought that we could consider $Y'$ any irreducible component of $Y$, and then look at the closed subset $f^{-1}(Y')$ of $X$. We can endow it with the reduced induced subscheme structure, and then look at the composition $f\circ j:f^{-1}(Y')\rightarrow Y$. Is it true that this map can be factorized through $Y'$? If so with which closed subscheme structure, the reduced induced one?
Is this the correct way of reducing the problem?
I thank you in advance for your precious help.