Exercise: pullback in the category of R-modules True or False:
(i) It always exist the pullback of two morphisms $f:A \rightarrow C$ and $g:B \rightarrow C$ in the category of R-modules.
(ii) if $f: A \rightarrow B$ is a homomorphism of R-modules then $(\text{Ker}(f), i,0)$ is the pullback of the morphisms $f: A \rightarrow B$ and $0:\{0\} \rightarrow B$ in the category of R-modules, where $i: \text{Ker}(f) \rightarrow A$ is the inclusion map and $\{0\}$ is the zero R-module.
Now, I've already proved the first item by showing that $(P,p,q)$ is the pullback of the two morphisms, with $$P=\left\{(a,b) \in A \times B:f(a)=g(b) \right\},$$ and $p: P \rightarrow A$, $q: P \rightarrow B$ defined as $p(a,b)=a$ and $q(a,b) = b$.
My question is, is the second item a direct consequence of the first one? I mean, if I have $f:A \rightarrow B$ and $0:\{0\} \rightarrow B$ then (with $A=A$ and $B=\{0\}$) by (i)
$$P=\left\{(a,b) \in A \times \{0\}:f(a)=0(b) \right\}=\left\{a \in A:f(a)=0_B \right\}= \text{Ker}(f).$$ Moreover $p(a,b)= p(a,0)=a$  is like $p(a) = a$ so in this case $p$ is the inclusion and $q(a,b) = q(a,0) = 0_B$ is always the projection in the second coordinate. So the pullback $(\text{Ker}(f), i,0)$ matches with the one established in the first item. Is this correct?
Thanks
 A: Well, the second item is not a direct consequence of the statement of the first item, which just says that some pullback exists.  But you are correct that it does follow easily from the specific pullback that you proved works for the first item, since the kernel described in the second item is almost a special case of that construction.  However, it is not literally a special case: your equation $$\left\{(a,b) \in A \times \{0\}:f(a)=0_B \right\}=\left\{a \in A:f(a)=0_B \right\}$$ is not actually true since $A\times\{0\}$ is not the same thing as $A$!  Of course, they are isomorphic by ignoring the second coordinate in $A\times\{0\}$, but you do have to do a little work to check that you do still have a pullback when you replace $A\times\{0\}$ with $A$ in this way.
(On a related note, I would strongly recommend to avoid using phrases like "the pullback" in situations like this unless you are sure you know what you are doing.  There is not really such a thing as "the pullback"; instead, there are many different pullbacks, which are isomorphic to each other but not literally the same.  Talking about "the pullback" instead of "a pullback" is a good way to get yourself confused if you are not well-versed in working with these concepts.)
