# Show that a homeomorphism between topological spaces $X, Y$ induces a homomorphism between the singular chain groups $C_n(X), C_n(Y)$

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.
• 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. – user35687 May 8 '20 at 16:12