Mathematics Asked by Falq on January 20, 2021
Let $C$ be a finitely complete category and $ain C$. Since $C$ is finitely complete, there exists (i.e. pullback) unique morphisms $m,n:atimes arightarrow a$ such that $mcirc1_a=ncirc 1_a$: i.e. $m=n$. We can take this $m$ to be our "multiplication". There is also a unique morphism $e:arightarrow1$ where $1$ is our terminal object; this can serve as our "multiplicative identity". How can I derive an $i:arightarrow a$ so that $(a,m,i,e)$ forms a group object in $C$?
I have used the fact that $C$ allows pullbacks to derive $m$. On the other hand, I have used the fact that $C$ possesses a terminal object to find $e$. What property gives me $i$?
As in the comments, you've messed up the direction of the unit; it's a map $e : 1 to a$, and such a map need not exist in a finitely complete category. An easy example is to take the category of $G$-sets for $G$ a nontrivial group; there a map $1 to X$ for a $G$-set $X$ corresponds to a fixed point, so we can take for example $X = G$ being acted on by left multiplication, which has no fixed points and hence no possible unit map.
In most categories with finite products (this is all you need to define group objects, and actually with a little fiddling you don't even need this), most objects don't have a group object structure; it's a quite special feature of $text{Set}$ that (assuming the axiom of choice) every set has a group structure. For example, in $text{Top}$ a group object is a topological group and the underlying topological space is heavily constrained, e.g. it must be homogeneous, its fundamental group must be abelian, etc.
Correct answer by Qiaochu Yuan on January 20, 2021
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