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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 "$S\xrightarrow{v}V,\, v=\{v_s\}_{s\in S}$".

enter image description here

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 $v$?

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    $\begingroup$ Just as an aside, never feel ashamed to ask a question. You are, after all, in the process of learning. $\endgroup$ – fosho Mar 15 '16 at 20:52
  • $\begingroup$ It means that paths having the same beginning and ending are equivalent. See Commutative diagram $\endgroup$ – Henricus V. Mar 15 '16 at 20:54
  • $\begingroup$ @Daniel I'm usually not, however, this was my first class in a completely new university, and I'm pretty sure I'm expected to know this, so I was afraid that this would be pretty badly received by my professor, and he's the one grading me after all, but yes, you're right. $\endgroup$ – YoTengoUnLCD Mar 15 '16 at 21:34
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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.

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  • $\begingroup$ It looks like I completely misunderstood what was being said: I thought $V$ was supposed to be some vector space of which we want $v$ to be a basis. But the set we want to be the basis is $V$? $\endgroup$ – YoTengoUnLCD Mar 15 '16 at 22:02
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    $\begingroup$ @YoTengoUnLCD : The set $S$ could be $\{1,\ldots,n\}$ and then for $s\in S$ you'd have $v(s) = v_s$, so $\{v(1)=v_1,\ldots,v(n)=v_n\}$ would be a basis of $V$. That makes $v$ a function whose domain is $S = \{1,\ldots,n\}$ and whose codomain is the vector space $V$. $\qquad$ $\endgroup$ – Michael Hardy Mar 15 '16 at 23:18
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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 $V=\{v_1,\ldots,v_n\}$.]

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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.

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  • $\begingroup$ Yes, I am familiar with the standard notions of basis and linear transformations. Thank you for your answer, I have a two questions: $1.$ In your answer, the 'symbol' which plays the role of a basis is $S$ while in Michael Hardy's, the basis would be $\{v(s):s\in S\}$, have I confused your answer? $2.$ I'm still confused as to how to understand this diagram, my current understand is something like that of a first order formula with a free variable, say $v$, and my goal is to be able to say that the diagram is satisfied if and only if $v$ is a basis (and we'd call $v$ a solution $\cdots$ $\endgroup$ – YoTengoUnLCD Mar 16 '16 at 0:25
  • $\begingroup$ of the diagram.). My last question in the main post is related to question $2$ in the above comment: If I understand the diagram correctly (as sort of a first order formula), then we can define (in the logical sense: en.wikipedia.org/wiki/Definable_set) the notion of basis using purely the language of categories. $\endgroup$ – YoTengoUnLCD Mar 16 '16 at 0:29
  • $\begingroup$ I know that what I just explained may be totally, utterly wrong, and I'm sorry for that, however, I'm trying to understand this stuff from some perspective I'm more familiar with. $\endgroup$ – YoTengoUnLCD Mar 16 '16 at 0:33
  • $\begingroup$ @YoTengoUnLCD, 1. Yes, mine is probably a misinterpretation, $S$ is an index set and $v$ is indexing function, but this is essentially the same as $S$ being subset itself. 2. You are thinking too much of diagrams, they have no meaning of their own. You still have to say: "For every $f$ there exists unique $g$ such that diagram commutes" to have a first order formula. Commuting diagrams are just convenient for expressing bunch of algebraic expressions. See for example five lemma. $\endgroup$ – Ennar Mar 16 '16 at 8:52

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