Relation between different definitions of homotopy

Question

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.

Answer 1

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.)

Answer 2

$\newcommand{\Z}{\mathbf{Z}}$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.