Functoriality of homology 
I am currently reading homological algebra and I am having trouble to understand where does the morphisms $im(d_{n+1}^{C})\longrightarrow{im}(d_{n+1}^{D})$ and $\ker(d_{n}^{C})\longrightarrow{\ker}(d_{n}^{D})$ come from.
My idea is: by definition the kernel is the equalizer $\ker(d_{n}^{D})\longrightarrow{D_{n}}\longrightarrow{D_{n-1}}$ and I need to show that $\ker(d_{n}^{C})\longrightarrow{C_{n}}\longrightarrow{D_{n}}\longrightarrow{D_{n-1}}$ is the zero morphism, this is done by using $d_{n+1}^{D}\circ{f_{n+1}}=f_{n}\circ{d_{n+1}^{C}}$.
Since $d_{n}^{C}\circ{\ker{d_{n}^{C}}}=0\iff{f_{n-1}}\circ{d_{n}^{C}}\circ{\ker{d_{n}^{C}}}=d_{n}^{D}\circ{f_{n}}\circ{ker{d_{n}^{C}}}=0$ by universality of the kernel, there exists a unique morphism $\ker(d_{n}^{C})\longrightarrow{\ker(d_{n}^{D})}$.
I have two questions:
i) Is my proof for the kernel correct or did I miss something?
ii) how can I show there exists a unique morphism from $im(d_{n+1}^{C})\longrightarrow{im(d_{n+1}^{D}})$?
 A: As statet in the comments, your line of thought for (1) is correct. For your second question, we can actually do the same thing again. Let us spell out the steps in (1) a little more so that we can see how this translates.
A commutative diagram illustrating the functorial properties of the kernel
The diagram depicts the universal property of the kernel in the way we want to use it. Since the right square commutes, $0 = f_{n-1}\circ d_n^C\circ \text{ker}(d_n^C) = d_n^D \circ f_n \circ \text{ker}(d_n^C)$. The universal property of the kernel now says that there is a unique morphism between the kernels, and the left square commutes.
To answer (2), we would actually need a definition of the image. Luckily, when working in an abelian category most of the image definitions tend to coincide. This is because we assume that there is an isomorphism $\overline{f} : \text{coIm}(f) \simeq \text{Im}(f)$, where the image is defined as the kernel of the cokernel of f, and the coimage is the cokernel of the kernel of f. This isomorphism is actually enough to give us image factorization.
With this in mind, we will work with the image as the kernel of the cokernel. This will give us a very similar diagram as before by first using the universal property of the cokernel, and then the universal property of the kernel.
A commutative diagram illustrating how we get the unique image morphism
There is a literature survey made by Theo Buhler on exact categories, which is a weaker variant of abelian categories, link to arXiv. The image factorization in exact categories is treated with much more care here.
