B Can the Diagram in the Article Be Interpreted as Commutative?

Stephen Tashi
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TL;DR
Does the given example of commutative diagram use conventional notation?
I'm used to seeing commutative diagrams where the vertices are mathematical objects and the edges (arrows) are mappings between them. Can the diagram ( from the interesting article https://people.reed.edu/~jerry/332/25jordan.pdf ) in the attached photo be interpreted that way?

In the article:

##k[x]## is the ring of polynomials over a field k.

##V## is a vector space.

##T## is a linear transformation on ##V##

CommDiagScreenshot.jpg


I understand ##T## and ##X## as maps, but do the vertical arrows go from a map to the argument of a map?
 
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The arrows (all of them, vertical and horizontal) go from element to element. Their LaTeX code is \mapsto.
 
I'll understand the vertical arrow on the left this way: There is an (unnamed) isomorphism mapping ##V## to a direct sum of quotient modules. So ##g(x) + <f_i(X)>## is one element of that direct sum and it is in the coset of the ##<f_i(X)>##. So the unnamed isomorphism maps an element of the direct sum to a vector ##v## in ##V##.
 
the visual clue to what fresh and Stephen are saying is that the arrows have little tails at the beginning which are perpendicular to the arrow.
 
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Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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