Define the image of a composition of maps

Question

I just asked myself the following question and couldn't see if or where there is an issue. Let $f\colon A\to B$ an arrow and let $i_{1}\colon A_{1}\rightarrowtail A$ and $i_{2}\colon A_{2}\rightarrowtail A$ two subobjects of $A$. Assume there is a map $k\colon A_{1}\to A_{2}$ not necessarily a monomorphism. The image of $A_{2}$ through the restriction $f\circ i_{2}$ (let's note it simply $I$ with $v\colon I\rightarrowtail B$) has the following property: it exists $\bar{g}\colon A_{2}\to I$ such $f\circ i_{2}=v\circ\bar{g}$. Composing with $k$, one gets $f\circ i_{2}\circ k=v\circ\bar{g}\circ k$. Now the image $J$ of $A_{1}$ through $f\circ i_{2}\circ k$ is such $f\circ i_{2}\circ k=w\circ\bar{h}$, with $w\colon J\rightarrowtail B$ and $\bar{h}\colon A_{1}\to J$. And, as we have just seen, $f\circ i_{2}\circ k=v\circ\bar{g}\circ k$ which means, by the universal property of the image of a map, that it exists a unique monomorphism $i\colon J\rightarrowtail I$ (making the whole thing commute). My (certainly naive) surprise is: why do I get an "inclusion" on the images of $A_{1}$ and $A_{2}$ while i didn't specify an inclusion on these two subobjects (there's no assumption for $k$ to be a monomorphism)? I would have expected something like with sets; i.e. $A_{1}\subseteq A_{2}\implies f\left(A_{1}\right)\subseteq f\left(A_{2}\right)$.