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.
$\DeclareMathOperator\im{im}\DeclareMathOperator\Ker{Ker}\require{AMScd}$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 $\im(gh)\leq\im\cp z{\ker f}$.