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For the free group the second arrow in the second diagram disappears.

Can someone give me an example of a definition of a universal object where the second arrow doesn't disappear? I have seen the definition of quotient space but unfortunately the second arrow disappears too.

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What do you mean by "it disappears"? – Martin Wanvik Apr 16 '12 at 15:17
@MartinWanvik I mean that you don't have to draw it because you can write $U(Y)$ instead of $Y$. – Rudy the Reindeer Apr 16 '12 at 15:19
As ineff's answer to the other question says, you shouldn't identify groups with their underlying sets. Groups and sets live in different categories (the category of groups and the category of sets respectively). To borrow from computer science, they are different types, and the forgetful functor typecasts between them. – Qiaochu Yuan Apr 16 '12 at 15:22
up vote 2 down vote accepted

Given your clarification above, here is an example that might satisfy you. Consider the category $\bf{Alg}$ of, say, complex unital algebras and the category $\bf{Grp}$ of groups. Define a functor $U: \bf{Alg} \to \bf{Grp}$ by taking each complex algebra $A$ to its group of units (i.e. invertible elements), and each algebra homomorphism $f: A \to B$ to its restriction to $U(A)$ (you can easily verify that this does indeed define a functor). Clearly, $U(A)$ and $A$ have different underlying sets, so this should satisfy your requirements.

This functor has a left adjoint (equivalently, there exists a universal morphism from every group to $U$), which associates to each group $G$ the (complex) group algebra $\mathbb{C}[G]$, defined as follows: the underlying vector space of $\mathbb{C}[G]$ is the space of finitely supported functions $f: G \to \mathbb{C}$, and the multiplication is $$ (f \ast g)(x) = \sum_{s,t \in G, st = x} f(s)g(t) = \sum_{s \in G} f(s)g(s^{-1}x), $$ for $f,g \in \mathbb{C}[G]$. The universal morphism $\Delta: G \to U(\mathbb{C}[G])$ is defined by $g \mapsto \delta_g$, where $\delta_g(s) = 1$ if $s = g$ and $\delta_g(s) = 0$ otherwise (you can easily show that this is a group homomorphism). A group homomorphism $\pi: G \to U(A)$, induces an algebra homomorphism $\hat{\pi}: \mathbb{C}[G] \to A$ by $$ \hat{\pi}(f) = \sum_{s \in G} f(s)\pi(s), $$ satisfying $\hat{\pi} \Delta = \pi$ (and it is the unique one with that property).

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Thank you very much! – Rudy the Reindeer Apr 19 '12 at 7:48

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