Definition of homomorphism of local rings

Let $(R,m_R)$, $(S,m_S)$ be two local rings. By definition, a local homomorphism of local rings is a ring homomorphism $f: R \to S$ such that the ideal generated by $f(m_R)$ in $S$ is contained in $m_S$: $f(m_R) \subseteq m_S$. According to Lemma 10.18.3, the condition $f(m_R) \subseteq m_S$ is equivalent to other conditions.

Question: What if $f(m_R)=m_S$? Does this imply something interesting, some stronger version of Lemma 10.18.3?

This question may be relevant, although I am not assuming thet $S$ is a finitely generated $R$-module.

Thank you very much!

Proposition: Let $R$ and $S$ are local rings with maximal ideals $\mathfrak{m}_R$ and $\mathfrak{m}_S$ and residue fields $k_R = R/\mathfrak{m}_R$ and $k_S = S/\mathfrak{m}_S$, respectively, let $\varphi : R\to S$ be a local ring homomorphism, and let $\overline{\varphi} : k_R\to k_S$ be the induced map on residue fields. Then the following are equivalent:

1. $\varphi$ is a local ring map such that $\varphi(\mathfrak{m}_R) = \mathfrak{m}_S$, and
2. If $s + \mathfrak{m}_S\in k_S$ is in the image of $\overline{\varphi},$ then $s$ is in the image of $\varphi.$

Proof: For any morphism $\varphi : R\to S$ of local rings, the snake lemma applied to $$ \require{AMScd} \begin{CD} 0 @>>> \mathfrak{m}_R @>>> R @>>> k_R @>>> 0\\ @VVV @V\left.\varphi\right|_{\mathfrak{m}_R}VV @V\varphi VV @V\overline{\varphi}VV @VVV\\ 0 @>>> \mathfrak{m}_S @>>> S @>>> k_S @>>> 0\\ \end{CD} $$ implies that we have a short exact sequence $$ 0\to\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R}\to\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}\to 0. $$ The map $\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R}\to\operatorname{Coker}\varphi$ is always injective, so that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ is an isomorphism if and only if $\operatorname{Coker}\left.\varphi\right|_{\mathfrak{m}_R} = 0.$ It follows that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ is an isomorphism if and only if $\left.\varphi\right|_{\mathfrak{m}_R}$ is surjective; i.e., if $\varphi(\mathfrak{m}_R) = \mathfrak{m}_S.$

Now, $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ being an isomorphism means that the class of $s + \mathfrak{m}_S$ in $\operatorname{Coker}\overline{\varphi}$ is $0$ if and only if the class of $s$ in $\operatorname{Coker}\varphi$ is. But the class of $s + \mathfrak{m}_S$ being $0$ is equivalent to $s + \mathfrak{m}_S$ being in the image of $\overline{\varphi},$ and similarly, the class of $s$ being $0$ is equivalent to $s$ being in the image of $\varphi.$

Observe that $s\in S$ being in the image of $\varphi$ always implies that $s + \mathfrak{m}_S$ is in the image of $\overline{\varphi}.$ Thus, it follows that $\operatorname{Coker}\varphi\to\operatorname{Coker}\overline{\varphi}$ being an isomorphism is equivalent to the condition that if $s + \mathfrak{m}_S\in\operatorname{Im}(\overline{\varphi}),$ then $s\in\operatorname{Im}(\varphi).$

Putting all of the above together, the proposition follows. $\square$

Loosely, the second condition of the proposition is similar to the fourth condition of lemma 10.18.3. The fourth condition in the lemma implies that units in $S$ lift to units in $R,$ and the second condition of the proposition implies that elements in the image of $\overline{\varphi}$ lift to elements in the image of $\varphi.$ However, observe that the liftability condition in the lemma is "vertical" in the sense of the commutative diagram in the proof of the proposition above, while the liftability condition in the proposition is "horizontal."