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When I did a course in topology, we defined "homotopy" in the following way:

"A homotopy between two continuous functions $f$ and $g$ from a topological space $X$ to a topological space $Y$ is defined to be a continuous function $H : X \times [0,1] \rightarrow Y$ from the product of the space $X$ with the unit interval $[0,1]$ to $Y$ such that, if $x \in X$ then $H(x,0) = f(x)$ and $H(x,1) = g(x)$. Two continuous functions are said to be homotopic if there exists a homotopy between them."

However, when I studied category theory, we defined it as follows:

"Let $A$ be an additive category and $\text{Ch}(A)$ the category of chain complexes of $A$. Let $X^{\cdot}$ and $Y^{\cdot}$ be two objects of $\text{Ch}(A)$, denoted by $$\cdots \overset{d_x^{n-2}}{\longrightarrow} X^{n-1}\overset{d_x^{n-1}}{\longrightarrow} X^n\overset{d_x^n}\longrightarrow X^{n+1}\overset{d_x^{n+1}}{\longrightarrow}\cdots $$ and $$\cdots \overset{d_y^{n-2}}{\longrightarrow} Y^{n-1}\overset{d_y^{n-1}}{\longrightarrow} Y^n\overset{d_y^n}\longrightarrow Y^{n+1}\overset{d_y^{n+1}}{\longrightarrow}\cdots $$ respectively. We say that a morphism $f\in\text{Hom}_{\text{Ch}(A)}(X^{\cdot},Y^{\cdot})$ is homotopic to zero if for every $n\in\mathbb{N}$ there exists a morphism $s^n:X^n\rightarrow Y^{n-1}$ such that $f^n=s^{n+1}\circ d_x^n+d_y^{n-1}\circ s^{n}$. The morphisms $s^n$ are called homotopies, and two morphisms $f$ and $g$ are said to be homotopic if $f-g$ is homotopic to zero."

I know there is a relation between them, but I can not see it. Of course the second definition does not make sense in topological spaces, but we could apply both definitions in the category of topological abelian groups.

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    $\begingroup$ A chain complex of abelian groups is the same thing as an abelian simplicial group under the Dold-Kan correspondence. A topological space is more or less the same thing as an abelian simplicial group under the adjunction that takes a space to its singular complex and a simplicial object to its geometric realization. Now check what happens to homotopic maps of spaces as you translate them to maps of chain complexes and vice versa. $\endgroup$ – Steven Landsburg Nov 10 '16 at 17:06
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    $\begingroup$ While Dold-Kan is a very good way to see that the definition of homotopy between chain complexes is a sensible one, it bears remember that a "homotopy" is in a sense part of the definition of your category. You might as well ask why the definition of maps of complexes and maps of topological spaces differ. $\endgroup$ – Denis Nardin Nov 10 '16 at 17:30
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    $\begingroup$ Steven: your `more-or-less the same' could be misleading as you added going to the abelian simplicial group and that transition destroys a lot of information, i.e. the difference between homotopy and homology. $\endgroup$ – Tim Porter Nov 11 '16 at 7:05
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Have a look at one of the standard texts such as Hatcher's book. Taking the singular complex of a topological space yields a simplicial set (or traditionally going one step further a chain complex). A topological homotopy yields a chain complex one between the induced maps. The difference between two homotopic chain maps will be homotopic to zero.

Another way is to show that there is a cylinder on the category of chain complexes so that mimicking the topological definition of homotopy gives a notion for chain maps which then can be shown to be equivalent to the one you give. (Again this can be found in many books on abstract ideas of homotopy.)

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  • $\begingroup$ A good place to start would be trace the relationship back to Quillen's abstraction of homotopy in his book 'Homotopical Algebra'. There are many more modern treatments of his notion of a Model Category. See for instance Baues's "Abstract Homotopy" or Riehl's "Categorical Homotopy Theory". The proof that the category or R-modules (every small abelian category embeds in the category of abelian groups by Freyd-Heron-Lubkin) can be found in Quillen or in "Homotopy Theories and Model Categories" by Dwyer & Spalinski. $\endgroup$ – Tyrone Nov 11 '16 at 15:36
  • $\begingroup$ Tyrone: the relationship is much older than Quillen and is very clearly explored in several of the really classical texts such as Mac Lane's Homology or Spanier's Algebraic Topology, without the clutter that model category theory surrounds it with. (The clutter is very useful for pushing things further but is not needed for the explicit question of the link between the two notions of homotopy.) My advice to the original questioner would be to keep the source simple (and possibly fairly classical) to start with. $\endgroup$ – Tim Porter Nov 11 '16 at 16:45
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    $\begingroup$ Actually the notion of cylinder functor predates Quillen by a few years and is in work of Dan Kan. (Quillen abstracted further of course.) This was then explored by Heiner Kamps before being picked up by Hans Baues. $\endgroup$ – Tim Porter Nov 11 '16 at 16:47
  • $\begingroup$ Quillen was actually heavily influenced by Kan's work if I recall and cites it as motivation. I just thought I would use it as an opportunity to advertise an open area of modern mathematics and some beautiful proofs if the original poster would like to delve a little deeper. And yes, I just noticed the initial of the second author of "Abstract Homotopy and Simple Homotopy Theory". $\endgroup$ – Tyrone Nov 11 '16 at 17:54
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$\newcommand{\Z}{\mathbf{Z}}$I'll talk about chain complexes over $\Z$ (although this works for any ring $R$). It looks like you're talking about cochain complexes in the question. Let $I$ be the analogue of the interval in chain complexes, namely the chain complex $\cdots\to 0\to \Z \to \Z\oplus\Z$, where the map $\Z\to\Z\oplus\Z$ sends $n\mapsto (n,-n)$. This is the singular chain complex of the interval $[0,1]$.

Let $C$ and $D$ be chain complexes over $\Z$, and let $f,g:C\to D$ be two chain maps. Define $\iota_C:C\to C\otimes I$ as: $(\iota_C)_n:C_n\to (C\otimes I)_n=C_n\oplus C_n\oplus C_{n-1}$ sends $x\mapsto (x,0,0)$, and similarly for $\iota_D$. A chain homotopy is a chain map $h:C\otimes I\to D$ such that $h\iota_C=f$ and $h\iota_D=g$. The requirement that this is a chain map is key.

The differential of $C\otimes I$ sends $d_n:x\otimes y\mapsto d^C_{|x|}x\otimes y + (-1)^{|x|}x\otimes d^I_{|y|}y$ where $|x|+|y|=n$. Therefore if $(x,y,z)\in C_n\oplus C_n\oplus C_{n-1}$, then $$d_n(x,y,z) = (d_n(x)+z,d_n(y)-z,-d_{n-1}(z))$$ The chain homotopy $h:C\otimes I\to D$ is degreewise $C_n\oplus C_n\oplus C_{n-1}\xrightarrow{(f_n,g_n,H_n)} D_n$. The requirement that $h$ be a chain map therefore means that (I'm afraid of indices getting messed up): $$d_n^D f_n(x)+d_n^D g_n(y)+d_n^D H_n(z) \\ = f_{n-1}d_{n-1}^C(x)+f_{n-1}(z) + g_{n-1}d_{n-1}^C(y)-g_{n-1}(z) - H_{n-1}d_{n-1}^C(z)$$ Because $f$ and $g$ are chain maps, this gives: $$f_{n-1} - g_{n-1} = d_n^D H_n + H_{n-1}d_{n-1}^C$$ which is the definition you know.

For a topological viewpoint on this, you can check that if $f\sim g:X\to Y$ where $X$ and $Y$ are topological spaces, then $f_\ast\sim g_\ast:S_\ast(X)\to S_\ast(Y)$ where $S_\ast(X)$ is the singular chain complex of $X$. A proof of this is in section 6 of my notes (http://www.mit.edu/~sanathd/18-905-notes.pdf) for Haynes Miller's algebraic topology class this fall at MIT.

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