Why is $f^{-1}f(U)=U+\ker f$? We are working in some abelian category. Given subobjects $Z, Z'\hookrightarrow X$ of $X$, we can define $Z\cap Z'= Z\times_X Z'$ as the pullback over $X$ and $Z+Z'= Z\amalg_{Z\cap Z'} Z'$ as the pushout over their intersection. We can also define the preimage of a subobject $U\hookrightarrow Y$ under a morphism $f\colon X\to Y$ as the pullback $f^{-1}(U)=X\times_Y U$.
Now, it is known (and trivial to see if one is willing to use elements) that $f^{-1}f(Z)=Z+\ker f$. However, I struggle to show this if I only want to use the above definitions of preimages and sums as well as the universal propery of kernels. Does anyone know a hint?
Edit: To account for Max' comment, we note that the inclusion morphism $Z+\ker f\to f^{-1}f(Z)$ can be constructed as follows. Since $Z\hookrightarrow X$ and $Z\to f(Z)$, we obtain $Z\to f^{-1}f(Z)$ by the universal property of $f^{-1}f(Z)$. Similarly, the maps $\ker f\hookrightarrow X$ and $0\colon\ker f\to f(Z)$ give us $\ker f\to f^{-1}f(Z)$; so by the universal property of $Z+\ker f$, we obtain a map $Z+\ker f\to f^{-1}f(Z)$, which, as pointed out by Max, must be a monomorphism since the composition $Z+\ker f\to f^{-1}f(Z)\to X$ is.
 A: $\DeclareMathOperator\im{im}\DeclareMathOperator\Ker{Ker}\require{AMScd}$Let consider the following commutative diagram of pullback squares:
\begin{CD}
P@>g>>Q@>h>>X\\
@VpVV@VqVV@VVfV\\
Z@>>z>X@>>f>Y
\end{CD}
and $\newcommand\cp[2]{\left[\genfrac{}{}{0}{}{#1}{#2}\right]}\cp z{\ker f}:Z\amalg\Ker f\to X$.
It's convenient to see $Z+\Ker f$ as the image of $\cp z{\ker f}$.
Since pulling back preserves image, it's enough to show
$$\im(gh)=\im\cp z{\ker f}$$
(diagrammatic composition order)
Since
$$\cp z{\ker f}f=\cp{zf}0=\cp 10zf$$
by universal property of pullback square we have $\cp z{\ker f}=agh$ for some morphism $a$, hence $\im\cp z{gh}\leqslant\im(gh)$.
On the other hand, let $k_1:Z\to Z\amalg\Ker f$ and $k_2:\Ker f\to Z\amalg\Ker f$ be the inclusions into the coproduct.
Then $(gh-pz)f=ghf-pzf=0$, hence $gh-pz\leq\ker f$.
Thus there exists a morphism $a$ such that $gh-pz=a\ker f$.
Let $b=pk_1+ak_2$.
Then
\begin{align}
b\cp z{\ker f}
&=(pk_1+ak_2)\cp z{\ker f}\\
&=pk_1\cp z{\ker f}+ak_2\cp z{\ker f}\\
&=pz+a\ker f\\
&=pz+gh-pz\\
&=gh
\end{align}
from which $\im(gh)\leq\im\cp z{\ker f}$.
