Let $D : \mathbf{I} \rightarrow \mathbf{A}$ be a diagram and $\left( L \xrightarrow{p_I} D(I) \right)_{I \in \mathbf{I}}$ a limit cone over $D$.

Theorem: If $A \xrightarrow{h,h'} L$ are maps such that $p_I \circ h = p_I \circ h'$ for all $I \in \mathbf{I}$, then $h = h'$.

Proof: Consider that $\left( A \xrightarrow{p_I \circ h} D(I) \right)_{I \in \mathbf{I}}$ is a cone over $D$. Then, there is a unique map $f : A \rightarrow L$ such that $p_I \circ f = p_I \circ h$. Taking $f = h$ and $f = h'$ both make this equality true (the former trivially and the latter by supposition), so we must have $f = h = h'$ since $f$ is unique.

$$\mjqed$$

Consider the case where $\mathbf{I}$ is the two-object discrete category, $\mathbf{A} = \mathbf{Set}$, and $A = \set{*}$. Interpreting the theorem above in this setting amounts to choosing a pair of sets $S, T$, and two elements of their product $L = S \times T$. The theorem then says that if two elements of $S \times T$ have the same left entry and right entry, then the two elements are equal as tuples -- in other words, equality of tuples is equivalent to component-wise equality.