proposition im(f) f span E in category theory

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)???

SOME REMARKS:

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''.

Thanks.

• Having a property about $h: U(X) \to U(Y)$ will usually not be enough. It is often required that $h$ is of the form $U(r)$ with $r: X\to Y$. Moreover the issue in your general setting is that $\operatorname{Im}(f)$ and $X$ live in different categories so that it's complicated to link the two (usually, "spanning" can be seen as "being an epimorphism") – Max Jul 19 '17 at 9:40
• Essentially, if you do the change I mentioned (i.e. change $h: U(X) \to U(Y)$ to $r: X\to Y$) you could say that the property you mention implies that, seeing $U$ as a "forgetful functor", any object whose "underlying set" contains $\operatorname{Im}(f)$ also contains $X$, that is, if $g$ is monic, then so is $h$. – Max Jul 19 '17 at 9:44
• I appreciate your comment @Max but I can't understand it unfurtenately. So, briefly, there is some statement in Category theory such as applied to Clifford algebras (tensor products and so on) recover the usual result Im$(i_E)$ spans $C_E$? Thanks again. – Dog_69 Jul 19 '17 at 16:20