If $A \subseteq B$ are ideals of $R$, can one define a surjective map from $R/B$ to $R/A$? If $A$ and $B$ are ideals of a commutative ring $R$ such that $A \subseteq B$, then can one define a surjective map from the quotient ring $R/B$ to $R/A$? I can define an injective map from $R/A$ to $R/B$, but not a surjective homomorphism from $R/B$ to $R/A$. Is it possible? I'm asking this because I have this sentence (which is true):
If $R$ is a UFD, then, if $h$ divides $t$ in R, the following sequence is exact (everything considered as a R-module)
$$ 0 \rightarrow hR/tR \rightarrow R/tR \rightarrow R/hR \rightarrow 0.$$ 
I just cannot see why there is a (natural?) surjective map $f: R/tR \rightarrow R/hR$.
 A: In general, if $A\subseteq B$, then $R/A$ will be "bigger" than $R/B$. For example, let $R = \mathbb{Z}, A = 4\mathbb{Z}, B=2\mathbb{Z}$. $R/A = \mathbb{Z}_4, R/B = \mathbb{Z}_2$, and there exists a surjective homomorphism from $R/A$ to $R/B$. So, I think what you want to ask is if we can find a surjective homomorphism from $R/A$ to $R/B$ (for the question you actually posed, the above is a counterexample). 
In fact, we can. Define $f: R/A\to R/B$ in the following manner. Let $r+A\in R/A$. Then there exists some $b\in R$ such that $r\in b+B$. Then let $f(r+A)=b+B$. 
$f$ is well-defined, for suppose $r+A=s+A$. Let $b_r, b_s\in R$ such that $r\in b_r+B, s\in b_s+B$. This is to say, $r=b_r+b_1, s=b_s+b_2$ where $b_1, b_2\in B$. Now, $r-s\in A$, and so $r-s\in B$. Further, $r-s=(b_r-b_s)+(b_1-b_2)$. It follows that $b_r-b_s\in B$, and so $b_r+B=b_s+B$, so $f$ is well-defined. 
$f$ is a homomorphism, for let $r,s\in R$. Then let $f(r+A)=b_r+B, f(s+A)=b_s+B$. Then $r+s \in (b_r+b_s)+B$, and so $f((r+s)+A) = (b_r+b_s)+B$. 
Finally, $f$ is surjective, for let $x+B\in R/B$. Then $f(x+A)=x+B$.
A: First, the natural map from $R/A$ to $R/B$ is surjective, and certainly not injective, unless $A=B$.
Second, the natural map $f\colon R/tR\longrightarrow R/hR$, $x+tR\longmapsto x+hR$ is well defined and surjective if $h\mid t$, because it translates into $tR\subseteq hR$, so we're in the previous case of $A\subseteq B$ (and this has nothing to see with factoriality).
A: I think one of the isomorphism theorems just gives
$(R/A)/(B/A)\cong R/B$ so that there is a surjective homomorphism from $R/A\rightarrow (R/A)/(B/A)\cong R/B$.
cf. https://en.wikipedia.org/wiki/Isomorphism_theorems#Rings
