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I'm trying to create a diagram for the countable direct limit in algebra. The following code

&&& A\ar@{-->}[dd]^{\exists!\alpha}\\
A_1\ar[r]^{\!\!\!\alpha_{1,2}} \ar@/^12pt/[urrr]|-{\alpha_{1}} \ar@/_12pt/[drrr]|-{\alpha'_{1}}&
A_2\ar[r]^{\!\!\!\alpha_{2,3}} \ar[urr ]|-{\alpha_{2}} \ar[drr ]|-{\alpha'_{2}}&
A_3\ar[r]^{\!\!\!\alpha_{3,4}} \ar[ur  ]|-{\alpha_{3}} \ar[dr  ]|-{\alpha'_{3}}&
&&& A'\\}$$

produces the diagram below:

enter image description here


  1. How can I make the arrows \alpha_1,\alpha_2,\alpha_3 arrive at their destination A?

  2. How to put labels \alpha_{1,2} lower (closer to their arrow)?

  3. How to move objects A and A' a bit to the right (along with \alpha_i,\alpha'_i) so that the arrow \alpha doesn't intersect \ldots and remains straight?

  4. When I bend arrows (using PDFtex) I a get rasterized arrow. How can I bend arrows and still have vector graphics?

  5. What is wrong with \alpha'_3?

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I would bend also the dashed arrow and also, slightly, the second diagonal arrows:

&&& A\ar@/^12pt/@{-->}[dd]^{\exists!\alpha}\\
\ar@/^4pt/[urr]|-{\alpha_{2}} \ar@/_4pt/[drr]|-{\alpha'_{2}}&
A_3\ar[r]^-{\alpha_{3,4}} \ar[ur]|-{\alpha_{3}} \ar[dr]|-{\alpha'_{3}}&
&&& A'

Note that \ar[r]^-{f} puts the label in the middle of the actual arrow.

enter image description here

A solution adding a column for the dots is not as good.

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Yeah, I've tried the option of bending the dashed arrow, but would still prefer moving A,A' to the right slightly. How about lowering \alpha_{n,n+1}? – Leon Sep 27 '11 at 19:41
This kind of things can surely be done, but it is always a pain to find something in the manual. :( Putting those labels under the arrow may avoid clashing in the middle. – egreg Sep 27 '11 at 20:24

Remove the spaces in the sqare brackets for \ar. This solves problems 1 and 5.

&&& A\ar@{-->}[dd]^{\exists!\alpha}\\
A_1\ar[r]^{\!\!\!\alpha_{1,2}} \ar@/^12pt/[urrr]|-{\alpha_{1}} \ar@/_12pt/[drrr]|-{\alpha'_{1}}&
A_2\ar[r]^{\!\!\!\alpha_{2,3}} \ar[urr]|-{\alpha_{2}} \ar[drr]|-{\alpha'_{2}}&
A_3\ar[r]^{\!\!\!\alpha_{3,4}} \ar[ur]|-{\alpha_{3}} \ar[dr]|-{\alpha'_{3}}&
&&& A'\\}

xy diagram

share|improve this answer
Ah, silly me, I forgot the space [... ] means something. That takes care of question 1), 5), thanks. What about the others? – Leon Sep 27 '11 at 18:32

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