I'm not sure if the title of the question is right, so there it goes an explanation.
Some algberaic structures such as tensor products or Clifford algebras for example, can be defined via a universal property.
In case of Clifford algebras, the pair $(C_E,i_E)$, wehere $C_E$ is a associative unitary $A$-algebra and $i_E$ is a Clifford map, is a Clifford algebra over $E$ if, for each associative unitary $A$-algebra, $\mathcal A$, and for each Clifford map, $\varphi:E \rightarrow \mathcal A$, there exists one algebra homomorphism (unique) $f:C_E$ such that $i_E\circ f = \varphi$ (the composition is done as ($f\circ g)(x) = g(f(x))$).
As a consequence, Im$(i_E)$ spans $C_E$. The same result is acchived for tensor products, exterior algebras and so on. So, this is why my question:
Let $\mathcal C,\mathcal D$ be two categories and let $U:\mathcal D \rightarrow \mathcal C$ be a functor from category $\mathcal D$ to $\mathcal C$. Consider $A$ an object of $\mathcal C$, $X$ an object of $\mathcal D$ and $f:A\rightarrow U(X)$ a morphism of Mor$(A,U(X))$. Suppose that, for each $Y\in$ Ob$(\mathcal D)$ and for each morphism $g:A\rightarrow U(Y)$ there exists a unique morphism $h:U(X)\rightarrow U(Y)$ such that $g=f\circ h$. Then, can I proved taht Im$(f)$ spans $X$ (in some way)???
i) My knowledge about Category theory is $\varepsilon$. I mean, I know the definition of category, object and morphism, and that's all.
ii) The universal property is copy from wikipedia's page: https://en.wikipedia.org/wiki/Universal_propert
iii) I have read something about the image of a morphism in Category theory. I don't undrstand the definition but I know is something more sofisticated than a set. Maybe this would be a problem for my guess and for that the add ''in some way''.