The Question: Let $f:C\rightarrow C'$ be a morphism. Let $j:I\rightarrow C'$ be the image of $f$. Suppose $\mathcal{C}$ is a category with equalizers, then the factorisation $f'$ of $f$ through its image is always epimorphic.
My attempt: Let $j:I\rightarrow C$ be the image of $f$ such that $j\circ f'=f$. To show that $f'$ is epimorphic we need to show for two morphisms $u,v:I\rightarrow T$ such that $u\circ f'=v\circ f'$ we have that $u=v$.
Let $e:E\rightarrow I$ be the equaliser of $u$ and $v$. Then we have that there is a unique $m:C\rightarrow E$ such that $e\circ m=f'$. From here I am stuck, I know I need to use the fact that $j:I\rightarrow C'$ is the image of $f:C\rightarrow C'$ but I can't figure out how...
Assuming your definition of image is initial factorisation through a monomorphism, notice that $f = (j \circ e) \circ m$ and hence by the definition of image there is a morphism $s : I \to E$ such that $j = j \circ e \circ s$. Now use that $j$ is mono to show that $e\circ s=1_I$ and hence $e$ is an isomorphism.