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In my Homological Algebra class we have defined and studied a lot of things through commutative diagrams. However, when we got to defining chain homotopies, we had to add a descriptive statement beyond "the diagram commutes". Wikipedia does the same thing, putting up this diagram and saying that we require $f-g = d_B h + h d_A$. Is there a way to read off this condition directly from the diagram, or to add information to the diagram directly so that the addition and subtraction can be read off of it?

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  • $\begingroup$ The equality of any two morphisms $\varphi$ and $\psi$ from $X$ to $Y$ is equivalent to the diagram $X \underset{\psi}{\overset{\varphi}{\rightrightarrows}} Y$ commuting, but maybe you're looking for something different? $\endgroup$ – Nate Gallup 2 days ago
  • $\begingroup$ @NateGallup yeah I think I'm looking for something different. I want to know if there's a way to read off from the diagram that $f^n - g^n = d_B^{n-1}h^n + h^{n+1}d_A^n$ for all $n$. In particular, I don't know how one would read the addition and subtraction off of the diagram $\endgroup$ – perpetuallyconfused yesterday
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    $\begingroup$ Note that the diagram that you drew in your question actually already does not commute (e.g. $f^n \neq g^n$), so putting additional arrows won't make it commute anyway. Addition and subtraction just add new arrows, i.e. $f^n - g^n$ and $d_B^{ n - 1} h^n + h^{n + 1} d_A^n$ are two new arrows from $A^n$ to $B^n$. My suggestion was that you have a separate diagram $A^n \underset{d_B^{ n - 1} h^n + h^{n + 1} d_A^n}{\overset{f^n - g^n}{\rightrightarrows}} B^n$ for these and require that this commutes in addition to the usual chain complex one. $\endgroup$ – Nate Gallup yesterday
  • $\begingroup$ @NateGallup I think that's what I'm looking for, would you be able to write one and put it as an answer? $\endgroup$ – perpetuallyconfused 16 hours ago

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