Ring Homomorphism Definition $\varphi:R\rightarrow S$ is said to be a ring homomorphism if, $R,S$ are rings and $\varphi$ is a map such that:
$\varphi(r_{1}+r_{2})= \varphi(r_{1})+\varphi(r_{2})$,
$\varphi(r_{1}.r_{2})= \varphi(r_{1}).\varphi(r_{2})$.
$\varphi(1_{R})= 1_{S}$
In the definition of a ring isomorphism, the third condition isn't stated, but here, we need the third condition. My question is, why do we need to state explicitly that the identity in $R$ is mapped to the identity in $S$. Can't we just deduce it from the definitions of a ring homomorphism?
EDIT: In reference to Anurag A's comment that rings are sometimes defined without the existence of unity, all the rings I am looking at will have unity in them, so if the answers are restricted to the case for when they do have unity, that would be extremely helpful
 A: No, $ℤ → ℤ × ℤ,~x ↦ (x,0)$ is not considered to be a ring homomorphism as it doesn’t preserve the identity element, yet is additive and multiplicative.
But an additive and multiplicative map of rings with identity elements $φ \colon R → S$ is at least guaranteed to be a ring homomorphism if it is surjective. That’s why you don’t need to require the identity elements to be preserved in the definition of a ring isomorphism – it’s automatically given.
This is because $\mathrm{img}~φ ⊂ S$ is a ring with identity $φ(1_R)$ as you can easily check for yourself. If $φ$ is surjective, it follows that $φ(1_R)$ is an identity for $S$ and from the uniqueness of identity elements $φ(1_R) = 1_S$.
A: The unit of a ring is unique, so if $\varphi: R\to S$ preserves addition and multiplication, and has an inverse, then it must send $1_R$ to $1_S$.
As others have pointed out, there is no similar argument if $\varphi$ is not assumed to have an inverse.  To take an extreme case, the zero map $R\to S$ satisfies the first two properties but not the third (unless $S=0$).
A: We cannot deduce that $\varphi(1_R)=1_S$ for a map $\varphi: R\to S$ that preserves addition and multiplication. The main obstruction is the existence of idempotents in $S$.
Indeed, $\varphi(1_R)=\varphi(1_R^2)=\varphi(1_R)^2$ and so $\varphi(1_R)$ is an idempotent in $S$.
If $S$ has no nontrivial idempotents, then this forces $\varphi(1_R)=0$ or $\varphi(1_R)=1_S$.
$\varphi(1_R)=0$ implies $\varphi(r)=0$ for all $r \in R$. Otherwise, 
$\varphi(1_R)=1_S$.
So all is well when $\varphi$ is not the zero map and $S$ has no nontrivial idempotents.
If $S$ is commutative and has a nontrivial idempotent $e$, then $\varphi(x)=ex$ is a map $S \to S$ that preserves addition and multiplication but does not send $1_S$ to $1_S$.
A typical example is $S=\mathbb Z \times \mathbb Z$ and $e=(1,0)$.
