Quick question about chain homotopies.

Question

In the definition of a chain homotopy (say $h$) between two chain maps (say $f$ and $g$), are the maps $h_i$ comprising the chain homotopy required to commute with all other maps involved (the $f_i$s, the $g_i$s, and the boundary maps in each complex)? Basically, in the associated diagram, are all the triangles commutative, or not necessarily? (I'd draw a picture, but I'm typing this on my phone, and that would take far too long). Thanks a lot for your help!

Answer 1

The short answer is that I would not expect that these triangles are commutative. Let us first reference the diagram and definition here. The triangles I think that you might be referring to the $(d^{n-1}_A,h^n,f^{n-1})$ and $(d^{n-1}_A,h^n,g^{n-1})$. Indeed if both triangles commuted, then this would mean that $f=g$. Likewise is true if the $(h^n,d^{n-1}_B,f^n)$ and $(h^n,d^{n-1}_B,g^n)$ all commuted.