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I wonder if someone can show me (or give me a sketch) why the following holds:

If A• is a subcomplex of a chain B•, then each chain in B• is homologous (in B•) to a chain in A•

Here B• = S• and A• is image of τ• of the functor S• where τ• is a transformator of S•

S• is singular chain complex.

Definition:

A transformator is a natural self-transformation τ• of the functor S• with τ0 = id.

Spelled out in detail this is, for each space X and each n ≥ 0, a homomorphism τn : Sn(X) → Sn(X) such that

(1) τ 0 = idS0(X) ,

(2) ∂ ◦ τ n = τ n−1 ◦ ∂,

(3) Sn(f) ◦ τ n = τ n ◦ Sn(f) for any map f : X → Y

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    One does not say *subchain complex* but *chain subcomplex*. – 2017-01-16
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    On the other hand, the claim you mention is false. Pick any non-exact chain complex $B$ and consider the zero subcomplex $A\subseteq B$. – 2017-01-16
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    You say that $B_*$ is equal to $S_*$ but do not tell us what $S_*$ is (Once you do tell us what $S_*$ is, notice that there is no real point in defining $S_*$ and then calling it $B_*$!). What $\tau$ is would probably be useful to make sense of what you wrote! – 2017-01-16
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    Sorry for being rush , here is edited form @MarianoSuárez-Álvarez Suárez-Álvarez – 2017-01-16
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    I was trying to pick an arbitrary chain in B• and present it as a finite linear combination (sum) of simplices in basis of Sn(X) and do the similar for chains in A• but I don't see how to present their difference as a boundary – 2017-01-16
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    Can you see how to do it in degree 0? (You can use LaTeX here, see [here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference); please do) – 2017-01-16
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    Well not actually since I only know then that I have to prove that the difference of any two chains in degree zero is boundary. – 2017-01-16
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    What is $A_0{}$? – 2017-01-16
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    The same as the zeroth singular chain group? – 2017-01-16
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    You are trying to prove that every element of $B_0$ is homologous to one in $A_0$, and you have just realized that $A_0$ is in fact equal to $B_0$. So: can you do the degree zero part of your claim? – 2017-01-16
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    No I cannot, because I cannot find specific element in dimension 1 which would be the boundary. I get that in this case my claim turns into showing that any two zero chains are homologous. – 2017-01-16
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    Again: you are trying to prove that every element of $B_0$ is homologous to one in $A_0$, and **you know that $A_0$ is equal to $B_0$**. Can you do the degree zero case? – 2017-01-16
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    But do I have to prove that it is homologous to just one chain (in which case is trivial) or to every chain in $A_0$ – 2017-01-16
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    Hm? Every element $x$ of $B_0$ is homologous to an element of $A_0$, namely to $x$ itself... – 2017-01-16
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    Yes, that's the trivial part, but I am confused since it's written that every chain in $B_0$ is homologous to a chain in $A_0$. The "a" article was confusing which is why I thought that I have to show homologus behaviour on each such chain in $A_0$ – 2017-01-16
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    That's quite an unreasonable reading of what you are trying to prove. It is moreover false, as you can check by considering what happens when you look at a discrete space with two points., – 2017-01-16
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    Ok, I get it now, thanks for making the zeroth part clear. But what about n$\geq$1 cases? Can I do it by induction or so – 2017-01-16
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    Before considering the case "$n\geq1$" I **emphatically** recommend that you do the case "$n=1$"… – 2017-01-16
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    So far, for the case $n=1$ I've got that the difference $T_1(c)$ - $c$ is a cycle for any chain c in $S_1(X)$ – 2017-01-16

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