Do two definitions for kernels match? Suppose that we work in Ab, the category of abelian groups. Consider a map $f : A \rightarrow B$ and let $\ker(f) = \{a \in A : f(a) = 0\}$. Now suppose that one can find a map $k : K \rightarrow A$ such that $fk = 0$ and for all maps $\eta : V \rightarrow A$ with $\eta f = 0$ there is a unique map $\theta : V \rightarrow K$ such that $\eta = k\theta$.
My question is: Do people consider $(k : K \rightarrow A) = (\ker(f) \subseteq A)$? I am not sure if there is an enlightening example for this. But usually the difference between $K$ and $\ker(f)$ are like the difference between $\{\pm 1\}$ and $\mathbb{Z}/(2)$.
Sometimes it is important to have equalities rather than isomorphisms, and it is very cumbersome to check which isomorphisms "work like equalities". I would like to say (loosely!) that any isomorphism from universal properties can be treated as equalities in the category that one works in (Ab in this case). In the case above, both $k$ and the inclusion $\ker(f) \hookrightarrow A$ share a common universal property, so there is a unique isomorphism between them (in a suitable category of arrows).
What I would like to hear is any justification (e.g. philosophical reason, specific examples) that says it is okay to consider canonical isomorphisms as equalities (or the other way around, if one may convince me).

Added (2/8/2014)
I apologize that I was not clear about what I meant by "canonical". When I say "two objects are canonically isomorphic" I mean that the objects have same categorical definition (i.e., they have a common universal property).
To me, it is less surprising that isomorphisms that are not from universal properties have some unexpected. We can consider a surjective abelian group map $A \rightarrow A$ with a nontrivial kernel $H$ (see Does $G\cong G/H$ imply that $H$ is trivial?), so in this case, we induce isomorphism $A/H \simeq A$.
Although this isomorphism is called "canonical", isomorphic objects of this kind may have some unwanted bizarre properties that they may have different torsion! I believe another example can be found in Patrick Da Silva's answer.
What I have been believing is that isomorphisms from universal properties do not have these unwanted behaviors, and the difference between two isomorphic objects (in categories of abelian groups, rings, modules over a fixed ring, vector spaces of a fixed field, etc) of this kind differ merely because they have "different names" in their elements, which is why I chose the example of $\{\pm 1\}$ and $\mathbb{Z}/(2)$ (one can also consider the set $\{0, 1\}$ with an obvious group structure to produce another example.) It was getting tiresome to prove that this kind of isomorphisms behave "as we expect" each time I encounter them, so I was wondering if there were any general facts or counterexamples that I was not aware of that corroborates or weakens my belief, respectively.
 A: The object $(k : K \to A)$ is uniquely defined up to isomorphism. But the set $\ker f$ together with the injection $i : \ker f \to A$ satisfies the property from $k : K \to A$, so if $(k,K)$ is a kernel for $A$ in $\mathrm{Ab}$, all we can say is that $K \simeq \ker f$ and that the obvious diagram that I cannot draw here commutes. 
I must say I am in general not a big fan of considering canonical isomorphisms as equalities, as in the case of the isomorphism $M/(M \cap N) \simeq (M+N)/N$. Even though the isomorphism is canonical, it can lead to some issues when trying to prove things ; for instance when one wants to prove that given an exact sequence of $R$-modules $0 \to N \to M \to P \to 0$ where $N$ and $M$ are graded and the map $N \to M$ is a graded homomorphism, one wants to show that $M/N \simeq P$ is graded ; saying that $(M_n + N)/N \simeq M_n/(M_n \cap N) = M_n / N_n$ perhaps brings faith, but doesn't prove anything. One must actually go through the details of showing that $\bigoplus_{n \ge 0} (M_n + N)/N$ is really a direct sum, since $A \cap B = 0$ does not imply $(A+N) \cap (B+N) = N$ (consider three generic lines as a subspace of $\mathbb R^2$ for example).
Hope that helps,
A: The subobjects of $A$ named by $k : K \to A$ and $\ker(f) \hookrightarrow A$ are indeed equal. However, I should note that a "subobject" is defined to be an isomorphism class of monic maps into $A$.
I should define what a morphism of "monic maps into $A$" is: if $g$ and $h$ are such things, then a morphism from $g$ to $h$ is a commutative diagram
$$ \begin{matrix}
\bullet &\xrightarrow{g}&  A
\\ \downarrow & & ||
\\ \bullet & \xrightarrow{h} & A
\end{matrix}$$
and composition of such morphisms amounts to composing the arrows on the left.

Another worthwhile point is how generalized elements work with subobjects. Recall that a generalized element of $A$ is interpreted as any morphism to $A$. A lot of the usual notions for elements make sense for generalized elements. e.g. if we have $f : A \to B$, and $x$ is a generalized element of $A$, then we can define the generlized element $f(x)$ to be $f \circ x$.
We can define a "membership" relation between generalized elements and subobjects: if $x$ is a generalized element of $A$, and $S$ is subobject of $A$, then $x \in S$ if and only if there is a commutative diagram
$$ \begin{matrix}
\bullet &\xrightarrow{x}&  A
\\ \downarrow & & ||
\\ \bullet & \xrightarrow{s} & A
\end{matrix}$$
where $s$ is any representative of the subobject $S$.
Note that if $x \in S$, then for any particular choice of $x$, the diagram above is uniquely determined, and each choice of $s$ gives an isomorphic diagram (where a morphism of such diagrams is defined in a suitable way), so the definition of $x \in S$ given is a reasonable one.
Note that in this language, given an arrow $f : A \to B$ between groups, there is a unique subobject $K$ of $A$ with the property that $x \in K$ if and only if $x \in A$ and $f(x) = 0$. (here, $0$ means the zero map with the same domain as $x$) So if we adjust to this categorical language, not only do we keep the simplicity of having a unique kernel, but we still retain the idea that there are many different monic arrows representing that kernel.
