Induced maps of the circle exponents Consider the family of circle self maps $\{f_n:S^1\to S^1:x\to x^n\}_{n\in\mathbb Z}$. How can we compute the induced maps $f_{n*}$ on the $1$-st simplical homology $H_1(S^1)$?
Thought $H_1(S^1)=\mathbb Z$ has for homomorphisms multiplication by an integer and no other homomorphisms. So $f_{n*}$ acts by multiplication with an integer. How can we see that it is multiplication with $n$?
Would it be useful to consider paths and their lifts to the universal cover? Could we consider the path $f(t)=e^{2\pi i t}$ and the composition $f_n \circ f$? I could not go far with this thought.
EDIT: I wanted to compute the induced maps on singular homology not simplical homology. Mr. Frenek resolved this issue.
 A: Intuitively, it just says that the image of $f_n$ goes $n$ times around the circle which is clear.
Let $T$ be a triangulation of the circle that consists of $3|n|$ edges, and let $S$ be a triangulation of the circle that consists of $3$ edges, assuming the usual orientation. Note that $f_n$ is homotopic to a simplicial map $g_n: T\to S$ that sends each of the $3|n|$ 1-simplices of $T$ to a $1$-simplex of $S$ in a consecutive way, "just around the circle" (both maps have degree $n$). The generator of $H_1(T)$ is the sum of all $3|n|$ simplices and this is sent to $n$ times the generator of $H_1(S)$ (which is the sum of all $3$ simplices).
Edit Based on the discussion, it seems to me that there is some confusion in your question: in simplicial homology, you need to have simplicial complexes and maps. True, the results are independent on the triangulation and any continuous maps between simplicial complexes can by approximated up to homotopy by a simplicial map between suitable subdivisions.
I'm not completely sure what are your definitions and sources, but my feeling is that the following idea might help you to unknot it. Let $T$ be a regular $3|n|$-gon and $S$ a triangle as above, and consider $h_1$ to be a homeomorphisms $T\to S^1$ that sends the first edge of $T$ to $S^1$ via $t\in [0,1]\mapsto e^{2\pi i t/(3|n|)}$ where the first edge of $T$ is linearly identified with $[0,1]$. Similarly, it maps the "second edge" of $T$ "linearly" to the circle via $t\mapsto e^{2\pi i (t+1)/(3|n|)}$ and so on. Similarly, choose a homeomorphism $h_2$ that maps $S$ to $S^1$ so that it maps the first edge $\simeq [0,1]$ via $s\mapsto e^{2\pi i s/3}$ and so on. Then the following diagram commutes (commutes exactly, not only up-to-homotopy, which would be also good enough).
$$
\begin{array}{ccc}
T & \stackrel{g_n}{\to} & S\\
\downarrow^{h_1} & &\downarrow^{h_2}\\
S^1 & \stackrel{f_n}{\to} & S^1
\end{array}
$$
The vertical arrows are orientation-preserving homeomorphism and take the generator of $H_1$ (in whatever sense) to generators of $H_1$ and the upper horizontal map is my simplicial map that "just multiplies by $n$" on the homology level, as shown above.
If your definitions of induced maps in simplicial homology are reasonable enough---whatever they are---you should be able to formalize this within your framework.
