First a few definitions:

Definition 1. The standard n-simplex is given by

$$\Delta^n = \{(t_0, t_1, \ldots , t_n) \in \mathbb{R}^{n+1} \vert\sum_{i=0}^{n} t_i = 1, t_i \geq 0, 0 \leq i \leq n \}.$$

Definition 2. A singular n-simplex in a topological space $X$ is a continuous map

$$\sigma\colon \Delta^n \rightarrow X.$$

Definition 3. A singular n-chain in $X$ is a finite formal linear combination $$\alpha = c_1\sigma_1 + c_2\sigma_2 + \cdots + c_m\sigma_m$$ with $c_i \in \mathbb{Z}$, $\sigma_i$ are singular n - simplices in $X$.

Let $C_n(X)$ be the group of all singular n-chains in $X$ with the natural addition:$$\alpha_1 + \alpha_2 := \sum_{i=1}^{m}(c_i+d_i)\sigma_i.$$

Let $X, Y$ be homeomorphic spaces. Let $f:X \rightarrow Y$ be a continuous map.

Question: According to the texts (e.g. Hatcher Algebraic Topology), we can define an induced homomorphism :

$$\tilde{f}:C_n(X) \rightarrow C_n(Y)$$

$$\tilde{f}(\sigma) = f\sigma$$

where for any singular n-simplex in $X$, $\sigma:\Delta^n \rightarrow X$, $f\sigma$ is a singular n-simplex in $Y$ $f\sigma:\Delta^n \rightarrow Y.$

For any linear combination $\Sigma_i a_i \sigma_i$ for $a_i \in \mathbb(Z), \sigma_i:\Delta^n \rightarrow X$,

$$\tilde{f}(\Sigma_i a_i \sigma_i) = \Sigma_i a_i \tilde{f}(\sigma_i) = \Sigma a_i f \sigma_i$$

How can we show that this is a homomorhism?

Here is what I have so far:

Let $\sigma_1$ and $\sigma_2$ be singular n-simplices in $X$.

Then, $$\tilde{f}(\sigma_1 \sigma_2) = f(\sigma_1 \sigma_2)$$

and $$\tilde{f}(\sigma_1) \tilde{f}(\sigma_2) = f(\sigma_1)f(\sigma_2)$$

How do we know these expressions are equal?

Edit: The notation in these expressions in not so accurate as the operation in the groups $C_n(X), C_n(Y)$ is $+$. See William's answer.

(Please give an answer in terms of general group theory and the things mentioned in this question; i.e. please no category theory.)


Any chain $c\in C_n(X;\mathbb{Z })$ is a sum

$$ \sum_{\sigma \in C(\Delta^n, X)} a_\sigma \sigma $$ where $C(\Delta^n, X)$ is the set of coninuous functions from the $n$-simplex to $X$, $a_\sigma \in \mathbb{Z}$, and $a_\sigma = 0$ for all but finitely many $\sigma$.

Then if $c_1 = \sum_\sigma a_\sigma \sigma$ and $c_2 = \sum_\sigma b_\sigma \sigma$ we have by definition

$$\begin{align} \tilde{f}(c_1) + \tilde{f}(c_2) &= \sum_{\sigma} a_\sigma f\sigma + \sum_{\sigma} b_\sigma f\sigma\\ &= \sum_\sigma (a_\sigma + b_\sigma) f\sigma \\&= \tilde{f}(\sum_\sigma (a_\sigma + b_\sigma)\sigma) \\&= \tilde{f}(c_1 + c_2) \end{align}$$

I know you said "no category theory" but this is actually just the universal property of the free product in the category of abelian groups. If $S$ is a set and $F(S)$ is the free abelian group generated by $S$, and $G$ is any abelian group, then any function $f\colon S \to G$ extends uniquely to a homomorphism $\tilde{f}\colon F(S) \to G$ whose formula is the same as what you've written down.

  • $\begingroup$ yes I have met that universal property; thanks for pointing it out. Question: when we show something is a homomorphism we need to show $f(xy) = f(x)f(y) \all x,y \in X$ it seems that you are only showing it for the specific $c_1, c_2$ in your answer i.e. you chose $c_1, c_2$ to not have any common $\sigma$ in their linear combination. Are you allowed to do that? what the general chain? Thanks. $\endgroup$ – user35687 May 8 '20 at 16:12
  • 2
    $\begingroup$ I changed my indices to make it a hopefully less confusing. $\endgroup$ – William May 8 '20 at 16:31

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