1
$\begingroup$

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.)

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.