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I have a small question:

Why is the following true?

"If we have a continuous mapping between two topological spaces $f:X\rightarrow Y$, we can associate a morphism of chain complexes $f_*\colon C_\bullet (X)\rightarrow C_\bullet(Y)$ "

thank you.

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closed as off-topic by Stefan Hamcke, Lord_Farin, Vedran Šego, Davide Giraudo, Trevor Wilson Oct 8 '13 at 19:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Stefan Hamcke, Lord_Farin, Vedran Šego, Davide Giraudo, Trevor Wilson
If this question can be reworded to fit the rules in the help center, please edit the question.

Make yourself clear what an element of $C_k(X)$ really is and how you "unavoidably" obtain an element of $C_k(Y)$ when also playing with $f$. – Hagen von Eitzen Oct 8 '13 at 16:49
i dont understand what you say @HagenvonEitzen – Vrouvrou Oct 8 '13 at 16:51
I've edited your question slightly. Please make sure I haven't altered the meaning in any way. – Dan Rust Oct 8 '13 at 19:49
up vote 4 down vote accepted

I suppose you are considering singular complexes for topological spaces: i.e. the complexes $(C_n(X),\delta_n)$ where $C_n(X)$ is the free $\mathbb Z$-module having as basis the set $\mathbf {Top}(\Delta^n,X)=\{\sigma \mid \sigma \colon \Delta^n \to X\}$ of continuous mappings from the standard $n$-simplex in the space.

If that's the case there's a very easy way to produce from a continuous mapping $f \colon X \to Y$ a chain map between the complexes $C_\bullet(X)$ and $C_\bullet(Y)$: you define the maps $f_n \colon C_n(X) \to C_n(Y)$ as those unique linear maps such that for every $n \in \mathbb N$ and for every $\sigma \colon \Delta^n \to X$ (which is an element of the basis of $C_n(X)$) you have that $f_n(\sigma)=f \circ \sigma$.

Of course you should verify that this homomorphisms $f_n$ are chain maps, i.e. commute with the $\delta$. But you can easily prove this by simply verifying the relation holds for the elements of the basis of the $C_n(X)$'s.

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