It seems reasonable for me that if $f:X\rightarrow Y$ is the constant map then $f_{*}:H_{n}(X)\rightarrow H_{n}(Y)$ is the zero map for $n>0$. But I don't see how to prove this. If $n$ is odd then it is OK as if $[p]_{n}$ denote the class of unique map from $\Delta^{n}\to Y$ then for odd $n$, $[p]_{n}=\partial[p]_{n+1}$. How to show for even $n$?
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I think it's simpler to see that if $f : X \to Y$ is a constant map, then it factors through as $X \to * \to Y$ where $*$ is a singleton. Therefore $f_*$ factors through $H_n(X) \to H_n(*) \to H_n(Y)$. But $H_n(*) = 0$ for $n > 0$, from which you conclude that $f_*$ is the zero map.
In fact this proof shows that any cycle in $C_n(X)$ (I guess it's singular homology here?) gets sent to zero on the level of chains for even $n$; because if $\sigma \in C_n(X)$ is a cycle, so is its image in $C_n(*)$, but there are no nontrivial cycles in $C_n(*)$ when $n>0$ is even. In particular it will be hard to find a chain whose boundary is $[p]_n$, because it's not even a cycle, actually!
PS: It's a general fact that working directly from the definition of the homology is way too complicated, except in very special cases and/or for very technical proofs. Working with properties of homology is much cleaner.
Najib Idrissi
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This seems logically suspicious ; considere the maps ${C_n(f) : C_n(X)\to C_n()}$ induced by $X\to $ and take any cycle $\sigma$ in $C_n(X)$, we have $\partial(f\circ\sigma)=\partial C_n(f)\sigma=C_{n-1}(f)\partial{\sigma}=0$. But also $f$ is trivial enough so that we can track $f\circ\sigma$, in fact it is just the map $\Delta_n\to$ and for $n$ even its boundary is not $0$ ; explicit computations leads to $\partial f\circ\sigma=\Delta_{n-1}\to$. How do we justify this contradiction ? – lou Jul 25 '22 at 19:15
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1@lou I think you confused cycles and simplices, since a single constant simplex will only be a cycle when it is in odd dimension, which is what you showed – bananananabatman Aug 11 '23 at 14:26