How do I read this diagram? I had my first category theory class today and the professor used these kind of diagrams, and terms like "the diagram commutes". I come from another university, and I have no idea what these kind of diagrams mean, and as the professor assumed everyone knew what they  mean, I was too ashamed to ask.
Next to the diagram, it should read "$S\xrightarrow{v}V,\, v=\{v_s\}_{s\in S}$".

Also, it should be "$\exists !$ linear transformation $g$ such that the diagram commutes".
The context is that he was trying to define what it means for a set to be a basis of a vector space in categorical language.
E: Thanks for the very enlightening answers, there are a few things that I still don't understand:


*

*If the only thing I'm given is the diagram (with the text I wrote) should I implicitly assume somehow that $S,V,W$ are sets and that some of them (I believe $V,W$) are vector spaces?

*What's the meaning of the diagonal $\equiv$ symbol in the centre of the diagram? Same question for the "broken" arrow from $V$ to $W$.

*My prof was treating this diagram as "an equation" where you are looking for a solution, this solution should be a basis of some vector space. Thus the diagram "defines" the notion of basis categorically, could someone explain this a bit more in detail? Which "element" is the one we're "solving for"? Is it little $v$? 
 A: Think of $v$ as a function whose domain is $S$ and whose codomain is $V$, and $f$ similarly as a function from $S$ to $W$.  It is saying there exists a function $g$ from $V$ to $W$ such that $g\circ v = f$.  The equality $g\circ v = f$, involving a composition of functions, is what it means to say this "commutes".
In a more abstract setting, $v$, $f$, and $g$ may be something other than function and the operation may be something other than composition of functions.
A: The sentence "the diagram commutes" means that for every $s\in S$, $f(s)=g(v(s))$ (that is, if you pick any of the possible directions from $s\in S$ you arrive to the same place in $W$)
Now, a good exercise for you would be to prove that if $S=\{1,2,\ldots,n\}$ $V=\{v_1,\ldots,v_n\}\subseteq \mathbb{R}$ is a base for $\mathbb{R}$ if and only if for every function $f:S\to W=\mathbb{R}$ there exists a unique function $g:V\to W$ such that the diagram commutes.
[Think of $g$ as a function that choose the coordinates of the vector $(f(1),\ldots,f(n))$ in the base $V=\{v_1,\ldots,v_n\}$.]
A: I'll assume you are familiar with standard notion of basis and that any linear operator can be defined just by specifying it's action on basis vectors and then extended by linearity. This is what this diagram describes:
Let $V$ be a vector space, and let $S$ be it's subset, and $v\colon S\to V$ set inclusion. Then, $S$ is basis of $V$ if and only if for any function $f\colon S\to W$ there exists unique linear map $g\colon V\to W$ such that $g\circ v = f$.
Now, this just means that $g$ is unique linear map that restricts to $f$. This is just the categorical statement of the above property that any function defined on basis can be uniquely extended by linearity.
Now, to your concrete questions:
Usually, you should assume that all objects, in this case $S$, $V$, $W$, lie in the same category, (e.g. category of sets, category of vector spaces). Your diagram is a bit simplified version of this kind of diagram, since every vector space is a set and every linear map is a function. The real story is happening in both category of sets and category of vector spaces: we are using a function $f$ between sets $S$ and $W$ to induce linear map $g$ between vector spaces $V$ and $W$. If you compare the linked diagram with your situation, left side would be inside category of sets, and right side in the category of vector spaces, while $U$ denotes that we "forget" that $V$ and $W$ are vector spaces and just remember that they are sets. I hope that this clarifies the confusion.
The $\equiv$ sign probably means that diagram commutes, i.e. $g\circ v= f$. Dashed line usually means induced map (morphism in the sense of category theory).
For the last part, I can't be sure what your professor really did mean, but you could think of this as a problem of defining linear map on whole vector space just by knowing its values on some subset. Being a basis is optimal solution to the problem since you can always uniquely extend any function defined on basis to linear map. All other subsets fail in that regard.
