**Proof.**
Proof of (1). We may work locally, hence we may assume that $Y = \mathop{\mathrm{Spec}}(A)$ and $X = V(f_1, \ldots , f_ c)$ where $f_1, \ldots , f_ c$ is a regular sequence in $A$. If $Z = \mathop{\mathrm{Spec}}(A/\mathfrak p)$, then we see that $f^{-1}(Z) = \mathop{\mathrm{Spec}}(A/\mathfrak p + (f_1, \ldots , f_ c))$. If $V$ is an irreducible component of $f^{-1}(Z)$, then we can choose a closed point $v \in V$ not contained in any other irreducible component of $f^{-1}(Z)$. Then

\[ \dim (Z) = \dim \mathcal{O}_{Z, v} \quad \text{and}\quad \dim (V) = \dim \mathcal{O}_{V, v} = \dim \mathcal{O}_{Z, v}/(f_1, \ldots , f_ c) \]

The first equality for example by Algebra, Lemma 10.116.1 and the second equality by our choice of closed point. The result now follows from the fact that dividing by one element in the maximal ideal decreases the dimension by at most $1$, see Algebra, Lemma 10.60.13.

Proof of (2). Choose a factorization as in the definition of a local complete intersection and apply (1). Some details omitted.
$\square$

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