Suppose I have a commutative diagram $D$ in the category of abelian groups. In $D$ there appear abelian groups $A$, $B$, $A\oplus B$ and some other abelian groups. Also, the natural inclusions $i:A\to A\oplus B$ and $j:B\to A\oplus B$ are also present in $D$. Suppose there is an abelain group $X$ in $D$ on which we have homolomorphisms $f:A\to X$ and $g:B\to X$ appearing in the diagram but there is, at present, no arrow from $A\oplus B$ to $X$ in $D$.
By the co product property of $A\oplus B$, there is a unique arrow from $h:A\oplus B\to X$ such that $h\circ i=f$ and $h\circ j=g$.
Question 1. Can adding $h$ to $D$ disturb the commutativity of $D$?
I think the the answer to the above question in NO.
Here is my reasoning. First we note that if $I$ is an initial object in a category $\mathcal C$, and $D$ is a commutative diagram in $\mathcal C$ which features $I$, then one can freely put an arrow starting at $I$ to any object $X$ in $D$ (if such an arrow is not already there) without disrupting the commutativity.
Now the direct sum of two abelian groups can be thought of as an initial object in the category of "cones" of $A$ and $B$. In other words, consider the category where the objects are triples $(X, f:A\to X, g:B\to X)$ and a morphism between $(X, f:A\to X, g:B\to X)$ and $(Y, h:A\to Y, k:B\to Y)$ is a homomorphism $F:X\to Y$ making the corresponding diagram commute.
Then by the previous remark about initial objects, we can answer the above question with a NO.
Question 2. Can we more generally say that if there is a universal object in a commutative diagram, then we can draw any missing arrows which can be drawn by using the universal property without disturbing the commutativity of the diagram.
Can somebody answer the above question?
The answer to Question 2 is no. Usually formally speaking a diagram in $\mathbb{C}$ is a functor into $\mathbb{C}$ using this as the definition one can make a simpler counter example, however some people might not feel that such an example constitutes a commutative diagram. I feel that nobody should object to the following: The diagram in the category of sets $$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{lll} \{*\}\ & \ra{\ } & \{*\} \\ \da{f} & & \da{f} \\ S & \ra{1_S} & S \end{array} $$ clearly commutes for any such $f$, however if $S$ is a set with at least two elements then inserting the diagonal arrow from $S \to \{*\}$ (which is the unique arrow to a terminal object) makes the diagram no longer commutative.
It should also be easy to show that answer to Question 1 is yes.