Fundamental groupoid Let $(X,x_0)$ be a pointed topological space. The homotopy groups $\pi_n(X,x_0)=Hom((S^n,s_0),(X,x_0))$ are groups because $S^n$ is a cogroup object in the pointed homotopy category.
Removing the basepoints leads to the fundamental groupoid: $\Pi_1(X)=Hom([0,1],X)$. Consequently $[0,1]$ must be a cogroupoid object in the category of topological spaces.
The problem is, I don't have a good enough grasp of what a groupoid object is in order to verify this directly. I saw that it can be treated as a special kind of category internal to the category of topological spaces, but I got bogged down in the details.
Questions:


*

*What are the axioms for a groupoid object, and why does $[0,1]$ satisfy those axioms in the category $Top^{op}$?

*Can one formulate a definition of groupoid object along the lines of group objects with only a partial multipication operation? If so, what would be the co-partial multiplication map $[0,1]\to [0,1]\sqcup [0,1]$?
 A: Perhaps one aspect of your difficulty with a cogroupoid object is that you are familiar with the usual algebraic structures, such as groups and rings, in which the algebraic operations are defined everywhere. However categories and groupoids are good examples of "partial algebraic structures". My definition of of higher dimensional algebra is that it is the study of partial algebraic structures whose algebraic operations are defined under geometric conditions. Thus the underlying geometric structure of a category is directed graph $s,t: C_1 \to C_0$ which is also reflexive in the sense that there is given a function $e: C_0 \to C_1$ such that $se=te$ is the identity on $C_0$.  You need the functions $s,t$ to decide when a multiplication of $f,g \in C_1$ is possible: it is defined on the pullback of $s,t$. 
A cocategory, or cogroupoid, again has an underlying  cograph, i.e. functions $s',t' :C_0 \to C_1$  but now the comultiplication is a function from $C_1$ to the pushout of $s',t'$. For an introduction see for example this presentation by Nigel Ray. ''
Thus a nice description of the unit interval $I$ is that it is a cogroupoid up to homotopy where homotopy is defined using the cograph structure on $I$. 
I mention that when I first started using groupoids I thought it was great to be able o get rid of base points; but I soon came round to the view  that what was needed was a set, say $A$, of base points and so to define the fundamental groupoid $\pi_1(X,A)$ on this set $A$. So for the circle $S^1$ it is useful to have $2$ base points. See also this mathoverflow question. 
Since you refer to higher homotopy group, you might find this mathoverflow question  relevant.  
December 27: In answer to your question in your comment below, you should download Higgins' Categories and Groupoids, and also compare with my book Topology and Groupoids. 
The standard books don't seem to like non path connected spaces, and  do not allow themselves the following construction: given a groupoid $G$ and a function $f: Ob(G) \to Y$, where $Y$ is a set, then form a new groupoid $U_f(G)$ with object set $Y$, and morphism $i: G \to U_f(G)$, having a nice universal property. The point is that making identifications on $Ob(G)$ allows compositions of elements of $G$ which were not possible before. So the groupoid $\mathcal I$ has a morphism $\iota:0 \to 1$ but when you identify $0$ and $1$ you get a new morphism $0 \to 0$ which you can compose repeatedly. This corresponds to making identifications of points in a space. (How do you get the circle from the unit interval $I$? Identify $0$ and $1$.) 
A more formal proof uses the universal properties of $\mathcal I$ and $\mathbb Z$ in the categories of groupoids and groups respectively. 
Having this construction $U_f$ enables  a single  account which yields free groups, free products of groups, and also the useful notion of free groupoid on a graph. 
A: The comultiplication is just the doubling map $t:[0,1]\to [0,1]\sqcup [0,1]/\sim$, where $\sim$ glues the first $1$ to the second $0$. Coassociativity is the fact that up to homotopy, the compositions $[0,1]\to ([0,1]\sqcup [0,1])\sqcup [0,1]/\sim$ and $[0,1]\to [0,1]\sqcup ([0,1]\sqcup [0,1])/\sim$ are equal. Hopefully this is reasonably clear as it stands: this seems to be a case in which thinking of a groupoid as a category, rather than a kind of algebra, makes things unnecessarily complicated.
