Assignment - Euler characteristic constant under barycentric subdivision I am working on an assignment and I am stuck, mostly I have no clue how to quite attack it. I don't want the answer or anything just advice on angels at which I can go about this. For an n-dimensional simplical complex $K$ do we have that the euler characteristic of it is
$$\chi(K) = \sum_{i=0}^n (-1)^i f_i $$
where $f_i$ is the number of i-dimensional simplices in $K$, the job is to show from this that $\chi(K)=\chi(K')$ from this definition. $K'$ is the first barycentric subdivision of $K$. The hint is to do it by induction on the number of simplices in $K$. This I feel doesn't help me because it requires one knows the quantity already from a formula which to my knowledge has not been supplied, Maybe I am missing something or am missunderstanding the question, but any hints and tips?
Again, I don't want any answers just hints for how to go about.
 A: There are more than one way to prove the invariance of Euler characteristic under barycentric subdivision. A self-contained and clear solution can be found here: http://www.math.uni-frankfurt.de/~johannso/SkriptAll/SkriptTopGen/SkriptTopI/vorl2.pdf . However, if the solution is to do induction on the number of simplicies, I believe the following solution is what you asked for. It does not make use of any lemma regarding Euler characteristic, but only some observation.
After barycentric subdivision, the Euler characteristic is transformed into $\chi(\text{sd}(K))=\sum_{n=0}^{N} (-1)^n \alpha'_n$ where $\alpha'_n$ is the number of $n$-simplices in sd($K$). We observe that a $n$-simplex is formed by, given a $(n-1)$-simplex and a $n$-th point outside the hyperplane generated by those $(n-1)$ vertices, joining the $n$-th point to all those $(n-1)$ vertices. Since any vertex of sd$(K)$ is a barycentre of some simplex of $K$, any $n$-simplex in sd$(K)$ can be interpreted as having the $n$-th vertex as the barycentre of some $i$-simplex in $K$, and the other $(n-1)$ vertices as forming a $(n-1)$-simplex in that $i$-th simplex. Denote $K^p_q$ as, after barycentric subdivision, the number of $q$-simplices (of sd$(K)$) in a $p$-simplex (of $K$) that does not have any vertice being the barycentre of the $p$-simplex. Note that $q\leq p$ is required and $K^p_p=0$. Then for any $n$, $0< n \leq N$:
\begin{align*}
\alpha_n'=\alpha_n K^n_{n-1} +\alpha_{n+1} K^{n+1}_{n-1} +...+\alpha_N K^N_{n-1}
\end{align*}
It is clear that $\alpha_0'=\alpha_0+\alpha_1+...+\alpha_N$. Then:
\begin{align*}
\chi(\text{sd}(K))
     &= \sum_{n=0}^{N} (-1)^n \alpha'_n\\
     &= \alpha_0 + \alpha_1(1-K^1_0) +\alpha_2(1-K^2_0+K^2_1)+...+\alpha_N(1-K^N_0+K^N_1-...\pm K^N_{N-1})\\
     &= C_0\alpha_0 + C_1\alpha_1 + C_2\alpha_2 +...+C_N\alpha_N
\end{align*}
where $\pm$ is $+$ if $N$ is even and $-$ if $N$ is odd. A quick check shows that $K^1_0=2$, so $C_1=1-K^1_0=-1=-C_0$. If we can show that for any $n\leq N$, $C_n=-C_{n-1}$, then $\chi'(\text{sd}(K))=\alpha_0-\alpha_1+\alpha_2-...\pm\alpha_N$ and we're done.
To do so, we carry out induction: with the hypothesis $K^{n+1}_q=(q+2)(K^n_q+K^n_{q-1})$ for $1\leq q\leq n-1\leq N-1$, we proceed to prove that $K^{n+1}_{q+1}=(q+3)(K^n_{q+1}+K^n_{q})$. Denote $k^r_s$ as the number of $s$-simplices in a $r$-simplex (before barycentric subdivision). By the same observation, the expression for $K_q^n$ is very similar to that of $\alpha_n'$ derived above:
\begin{align*}
K^n_q = k^n_qK^q_{q-1} + k^n_{q+1}K^{q+1}_{q-1}+...+k^n_{n-1}K^{n-1}_{q-1}
\end{align*}
Also, we note that $k^{n+1}_q=k^n_q+k^n_{q-1}$, and $k^n_n=1$. Plug in this relation and the induction hypothesis, a few steps of simple algebraic manipulation obtain the conclusion. By the same procedure, we can check that (using $K^{n+1}_0=2(K^n_0+1)$) $K^{n+1}_1=3(K^n_1+K^n_0)$, which can serve as the starting point for induction. Therefore $K^{n+1}_q=(q+2)(K^n_q+K^n_{q-1})$ for $1\leq q\leq n\leq N$. 
Now, for $n$ even:
\begin{align*}
C_n &=1-K^n_0 + K^n_1 - K^n_2 +...-K^n_{n-2}+K^n_{n-1}\\
    &= 1 - 2(K^{n-1}_0+1) + 3(K^{n-1}_1 + K^{n-1}_0) - 4(K_2^{n-1}+K_1^{n-1} \\+ ... &-n(K^{n-1}_{n-2}+K^{n-1}_{n-3}) + (n+1)(K^{n-1}_{n-1}+K^{n-1}_{n-2})\\
    &= -1 + K^{n-1}_0 - K^{n-1}_1 + ... + K^{n-1}_{n-2}\\
    &= -C_{n-1}
\end{align*}
The same result yields for $n$ odd, by explicit calculation like above. Therefore, the Euler characteristic is invariant under barycentric subdivision.
A: 1) The writer of the assignment probably had the following in mind: first note that Euler characteristic X is a valuation $X(A \cup B) = X(A) + X(B) - X(A \cap B)$ by checking first for $f_i$ by counting  then extend by linearity. Now first verify the invariance for a simplex S by induction on dimension. When adding a simplex S to a complex K it must intersect in a union of smaller dimensional simplexes. Now, since by induction assumption, S and K as well as $A \cap B$ are invariant under subdivision  by the above inclusion-exclusion formula, also the larger complex $K \cup S$ does not change Euler characteristic when applying a Barycentric subdivision.
2) The cheapest proof is probably to look at the topological realization $|K|$ of the simplicial complex. Since that realization has the same Euler characteristic than $K$ and $|K|=|K'|$ the invariance is obvious.   
3) There is an other simple combinatorial proof using the notion  of homotopy for simplicial complexes, a notion developed by  Whitehead for simplicial complexes and translated into the language of  graph theory by Ivashchenko and simplified by Chen-Yau-Yeh.  Its quite easy to show that Barycentric subdivisions are homotopies  with this definition and that elementary homotopy steps preserve the  Euler characteristic.   
4) The cohomology proof mentioned before is also relatively simple. Since Euler-Poincare shows that the Euler characteristic is  equal to cohomological Euler characteristic (a computation in  the discrete first done by Benno Eckmann), one can first check that cohomology is invariant under subdivision (by extending cocycles and coboundaries) and so get again the invariance.   
5) Here is a cool proof: the f-vector f transforms like f'=Af where A(i,j) = i! S(j,i) where S(j,i) are the Stirling numbers of the second type. I had myself fought for a while with finding a formula for the change of the f-vectors: See here, where also a proof of the recursion can be found: the argument is that every (k+1)-simplex gets split into (k+1)! subsimplices under subdivision. This formula could be shown by induction. But there are new (k+1)-simplices spawn off from higher dimensional simplices. This is why the matrix A is upper triangular with diagonal entries (k+1)!  The magic is that the vector a=(1,-1,1,-1 ...) is an eigenvector of the transpose of A to the eigenvalue 1. This means that X(K) = (a,f) = (AT a,f) = (a,A f) = (a,f')=X(K') showing that the Euler characteristic is constant. [By the way, the other eigenvectors are exciting too as they lead only to invariants like the Euler characteristic, if they are zero. If the simplicial complex comes from a triangulation of a d-manifold and k+d is even, then this new Barycentric characteristic is zero. They could be deduced also from Dehn-Sommerville-Klee identities but the Barycentric approach gives more information. See  this document for lots of references.]     
A: I think I have an elementary proof.
Suppose that $K$ is a simplicial complex, where
$$
\chi(K)=\sum_{i=0}^N(-1)^if_i.
$$
Assume for each $n$-simplex $\Delta^n$, the number of inner $i$-simplices in $\mathrm{sd}(\Delta^n)$ is $f_i^n$. We only consider inner $i$ simplices. This is because when we count the number of simplices in $\mathrm{sd}(K)$, the $i$-simplicies can only come from the barycentric subdivision of an $n$-simplex where $n\geq i$, and by counting inner simplicies, no simplex would be counted twice. So the number of $i$-simplicies in $\mathrm{sd}(K)$ should be the sum of $f_nf^n_i$ where $n$ takes all possiblities $\geq i$. Then
$$
\chi(\mathrm{sd}(K))=\sum_{i=0}^N(-1)^i(\sum_{n=i}^Nf_nf^n_i)\\
=\sum_{n=0}^Nf_n(\sum_{i=0}^n(-1)^if^n_i).
$$
Hence it suffices to prove that
$$
\sum_{i=0}^n(-1)^if^n_i=(-1)^n.
$$
We shall use induction to prove this. Obviously $f^n_0=1$. A simplex in $\mathrm{sd}(\Delta^n)$ can be viewed as a chain of sets of bartcentres, for instance
$$
[0,1,\cdots,n]\supset[0,1,\cdots,n-1]\supset\cdots\supset[0,1]\supset[0]
$$
is an $n$-simplex in $\mathrm{sd}(\Delta^n)$. A simplex is inner if and only if the first barycentre is $[0,1,\cdots,n]$, which is the barycentre of $\Delta^n$. Given an $i$-simplex in $\mathrm{sd}(\Delta^n)$, if we delete the first barycentre, then it will become an inner $i-1$-simplex of some $d$-simplex, where $i-1\leq d\leq n-1$, and for a given $d$, we can delete $n-d$ vertices, which gives ${n+1}\choose{n-d}$ possiblities, therefore
$$
f^n_i=\sum_{k=1}^{n-i+1} {{n+1}\choose{k}} f_{i-1}^{n-k}.
$$
Notice that
$$
\sum_{i=0}^n(-1)^if^n_i=f^n_0+\sum_{i=1}^n(-1)^if^n_i\\
=f^n_0+\sum_{i=1}^n(-1)^i\left(\sum_{k=1}^{n-i+1} {{n+1}\choose{k}} f_{i-1}^{n-k}\right)\\
=f^n_0+\sum_{k=1}^n{{n+1}\choose{k}}\left(\sum_{i=1}^{n-k+1}(-1)^if_{i-1}^{n-k}\right).
$$
By assumption,
$$
\sum_{i=1}^{n-k+1}(-1)^if_{i-1}^{n-k}=\sum_{i=0}^{n-k}(-1)^{i+1}f_{i}^{n-k}=(-1)\sum_{i=0}^{n-k}(-1)^if_{i}^{n-k}=(-1)^{n-k+1},
$$
hence
$$
f^n_0+\sum_{k=1}^n{{n+1}\choose{k}}\left(\sum_{i=1}^{n-k+1}(-1)^if_{i-1}^{n-k}\right)\\
=f^n_0+\sum_{k=1}^n(-1)^{n-k+1}{{n+1}\choose{k}}\\
=f^n_0+(-1)^{n+1}\sum_{k=1}^n(-1)^k{{n+1}\choose{k}}\\
=1+(-1)^{n+1}\left((1-1)^{n+1}-(-1)^0{{n+1}\choose{0}}-(-1)^{n+1}{{n+1}\choose{n+1}}\right)\\
=\left\{\begin{matrix}1&n\text{ is even}\\ -1&n\text{ is odd}\end{matrix}\right..
$$
