# Morphisms of complexes chain [closed]

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 WilsonOct 8 '13 at 19:01

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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. –  Daniel Rust Oct 8 '13 at 19:49

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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