Suppose $\textsf{T}_i :\textsf{V}_i \to \textsf{V}_{i+1}$ ($i=  0,\dots,n$) is a sequence of vector spaces such that  $$\ker( \textsf{T}_{i+1} ) = \operatorname{im}( \textsf{T}_i )$$ with $\textsf{V}_0 = \textsf{V}_{n+1} = \{0\}$. Show that $$\sum_{i=1}^n (-1)^i \dim (\textsf{V}_i ) = 0$$
Q) Prove that if $0\rightarrow V_1\rightarrow V_2\rightarrow...\rightarrow V_k\rightarrow 0$ is an exact sequence of finite dimensional vector spaces, then $\sum_{i=1}^{k}(-1)^idim (V_i) = 0$.  For $k=3$, I see that the statement is true. For $k&gt;3$, am I correct in understanding that $V_2 = V_1\oplus V_3 (\implies dim(V_2)=dim(V_1)+dim(V_3))$ and $V_3 = V_2\oplus V_4(\implies dim(V_3)=dim(V_2)+dim(V_4))$ and so on?   If it is true, then for $k=4$, $\sum_{i=1}^{4}(-1)^idim (V_i) = -m_1+m_1+m_3-m_2-m_4+m_4$ and am not sure why $m_2=m_3$ for the sum to be $0$? where $m_i = dim(V_i)$. Thanks.
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