This is just a naive comment on the second to last paragraph in Qiaochu's answer:

'So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.'

Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? Of course this breaks down when we no longer restrict to vector spaces but from the point of view of working over a field the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural.

The signs in the formula are then intepreted as arising from the braiding i.e. Given complexes $(C,\partial)$ and $(D,\delta)$, we evaluate $\partial \otimes 1 + 1 \otimes \delta$ on tensors $\sum_i c_i \otimes d_i$ via the pairing

$$\mathrm{End}(C) \otimes \mathrm{End}(D) \otimes C \otimes D \xrightarrow{\Psi} \mathrm{End}_k(C) \otimes C \otimes \mathrm{End}(D) \otimes D \to C \otimes D $$

Where $\Psi$ is the braiding morphism. We see that taking $\Psi$ to be the plain tensor flip doesn't work and then modify this so that on homogeneous elements $c$ and $d$ we have $\Psi\colon c \otimes d \mapsto (-1)^{|c||d|} d \otimes c$ which does work.

To me this is main unsatisfactory part of the story: why the Koszul braiding? The only explanation I have seen so far is: 'It works! Just accept it.' I guess if I could see that the Koszul braiding was the only braiding that worked then I would be happy but I can't see that off the top of my head...

This is just a naive comment on the second to last paragraph in Qiaochu's answer:

'So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.'

Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? From this point of view the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural.

The signs in the formula are then intepreted as arising from the braiding i.e. Given complexes $(C,\partial)$ and $(D,\delta)$, we evaluate $\partial \otimes 1 + 1 \otimes \delta$ on tensors $\sum_i c_i \otimes d_i$ via the pairing

$$\mathrm{End}(C) \otimes \mathrm{End}(D) \otimes C \otimes D \xrightarrow{\Psi} \mathrm{End}_k(C) \otimes C \otimes \mathrm{End}(D) \otimes D \to C \otimes D $$

Where $\Psi$ is the braiding morphism. We see that taking $\Psi$ to be the plain tensor flip doesn't work and then modify this so that on homogeneous elements $c$ and $d$ we have $\Psi\colon c \otimes d \mapsto (-1)^{|c||d|} d \otimes c$ which does work.

To me this is main unsatisfactory part of the story: why the Koszul braiding? The only explanation I have seen so far is: 'It works! Just accept it.' I guess if I could see that the Koszul braiding was the only braiding that worked then I would be happy but I can't see that off the top of my head...

This is just a naive comment on the second to last paragraph in Qiaochu's answer:

'So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.'

Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? Of course this breaks down when we no longer restrict to vector spaces but from the point of view of working over a field the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural. We are still left with the problem of why that particular choice of braiding but maybe we could just say that it is the 'simplest' modification of the plain flip that works.

This is just a naive comment on the second to last paragraph in Qiaochu's answer:

'So what you've observed is that the category of chain complexes is morally closer to the second construction than the first. Why this is the case is something that I don't have a completely satisfactory answer for, but in any case the point I want to make is that the second comultiplication is just as natural as the first from the appropriate perspective.'

Surely the motivation comes from thinking of the associated double complex and the desire to 'embed' both complexes into the tensor product complex? Of course this breaks down when we no longer restrict to vector spaces but from the point of view of working over a field the choice $x \mapsto x \otimes 1 + 1 \otimes x$ is very natural.

The signs in the formula are then intepreted as arising from the braiding i.e. Given complexes $(C,\partial)$ and $(D,\delta)$, we evaluate $\partial \otimes 1 + 1 \otimes \delta$ on tensors $\sum_i c_i \otimes d_i$ via the pairing

$$\mathrm{End}(C) \otimes \mathrm{End}(D) \otimes C \otimes D \xrightarrow{\Psi} \mathrm{End}_k(C) \otimes C \otimes \mathrm{End}(D) \otimes D \to C \otimes D $$

Where $\Psi$ is the braiding morphism. We see that taking $\Psi$ to be the plain tensor flip doesn't work and then modify this so that on homogeneous elements $c$ and $d$ we have $\Psi\colon c \otimes d \mapsto (-1)^{|c||d|} d \otimes c$ which does work.

To me this is main unsatisfactory part of the story: why the Koszul braiding? The only explanation I have seen so far is: 'It works! Just accept it.' I guess if I could see that the Koszul braiding was the only braiding that worked then I would be happy but I can't see that off the top of my head...