Prove that for a path connected topological group $G$ the Euler characteristic $\chi (G)$ is zero. Prove that for a path connected topological group $G$ the Euler characteristic $\chi (G)$ is zero.
I'm trying to use Lefschetz for this proof. Bare with me. 
The Lefschetz Fixed Point Theorem states that: 
If $X$ is a finite simplical complex, or more generally a retract of a finite simplicial complex, and $f:X\longrightarrow X$ is a map with $\tau (f) \neq 0$, then $f$ has a fixed point.  Here, the Lefschetz Number $\tau (f)$ is defined to be $\tau (f) = \Sigma _n(-1)^ntr(f_{*}:H_n(X)\longrightarrow H_n(X))$.
Here is a blurb explaining the proof: "The Lefschetz fixed point theorem tells us that if an endomorphism on a space has Lefschetz number (the alternating sum of the traces of its induced maps on homology) not equal to zero, then it must have a fixed point. Groups famously have lots of topological automorphisms with no fixed points (left-multiply by any nontrivial element g). One can also prove that such a function is homotopic to the identity whenever the group is path connected (just take any path from 1 to g and appropriate translations thereof performed simultaneously). Hence, since homotopic maps induce the same maps on homology, the Lefschetz number of the identity map must be zero, and the Lefschetz number of the identity map is precisely the Euler characteristic!"
But I'm a little confused.  When they say "such a function is homotopic to the identity" are they referring to the function $f: X\rightarrow X$? I feel like they mean homotopic to the constant map. I know $X$ is path-connected, but I thought if $X$ was contractible, then it was homotopic to a constant map. 
Moving past that, so if $f$ is homotopic to the constant map, then we would also have that $f_{*}$ is homotopic to a constant map.  Which would leave $\tau (f) = 0$. Since $f$ is the identity map on $X$, then $\tau (f) = \chi (X)$.  So, $\chi (X) = 0$.
Can someone help clarify this explanation and/or give an alternative way to prove this.  
Also, I'm not given that $X$ is a simplicial complex, so I might be in the wrong with even using Lefschetz.  All we're given about $X$ is that it is path connected topological group.  
 A: They mean homotopic to the identity. The constant map induces zeros on homology - it is the map that sends everything to a point, and the identity is homotopic to the constant map iff the space is contractible (by definition). 
The identity map induces the identity on homology, so the trace gives the dimension of the vector space. (Trace of the identity matrix.)
Hence the Lefschetz number for the identity map is the alternating sums of the dimensions of the homology groups, i.e. the Euler characteristic. But then because the identity is homotopic to a right translation by some $g \not = e$, and the Lefschetz number is invariant under homotopies (because homotopic maps induce the same map on homology), and because $R_g$ has no fixed points, the lefschetz number must be zero.
(As JasonDeVito points out - you need compactness to apply Lefschetz.)
This is a cool argument - thanks for sharing it. Let me know if you have more questions.
I don't know the answer to your question about simplicial complexes, though I wouldn't worry about it too much - I think most spaces you encounter can be given that structure.
