Let $F=F(H,H)$ be the space of bounded Fredholm operators in a Hilbert space $H$ with topology inherited from the norm operator topology, and let $X$ be a compact topological space.


For a continuous map $T\colon X\to F$, there exists a closed subspace $W\subseteq H$ with $\dim H/W<\infty$ such that $W\cap\ker T\_x=0$ for all $x\in X$ and $H/T(W) =\bigcup\_{x\in X} H/T\_x(W)$ is a vector bundle over $X$ (See appendix of *K-Theory, Anderson & Atiyah*).
Then one can show that
$$\mbox{Ind}\_1(T) = [X\times H/W] - [H/T(W)] \in K(X)$$
does not depend on $W$.


On the other hand, there exists a finite dimensional subspace $V\subseteq H$ such that $V+T\_x(H) = H$ for all $x\in X$, so we can define $T^V\colon X\to F(H\oplus V, H)$ by $T^V\_x(u,v) = T\_x u + v$. Then $T^V\_x$ is surjective and $\dim\ker T^V\_x$ is constant on $x$. Thus $\ker T^V = \bigcup\_{x\in X} \ker T\_x$ is also a vector bundle over $X$. One can show that
$$ \mbox{Ind}\_2(T) = [\ker T^V] - [X\times V] \in K(X)$$
does not depend on $V$.


These index maps are called *the* family index of families of Fredholm operators in $H$, and it made me suspect that they are equal.



> 
> **Question:** Is it true that $$[X\times H/W] - [H/T(W)] = [\ker T^V] - [X\times V] \qquad (1)$$ in $K(X)$ ?
> Is there any reference that proves the equivalence of these indexes?
> 
> 
> 




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Edit: We can shrink $W$ or augment $V$ in order to have $\dim H/W = \dim V$. Say $H/W \cong V \cong \mathbb{C}^N$, so that $X\times H/W \cong X\times V \cong X\times\mathbb{C}^N$, and therefore
$$\mbox{Ind}\_1(T) = [X\times\mathbb{C}^N] - [H/T(W)]\ ,$$
$$\mbox{Ind}\_2(T) = [\ker T^V] - [X\times\mathbb{C}^N]\ .$$


Equation $(1)$ becomes
$$[X\times\mathbb{C}^N] - [H/T(W)] = [\ker T^V] - [X\times\mathbb{C}^N]$$
and it holds iff there exists $k\geq0$ such that


$$\ker T^V \oplus H/T(W) \oplus (X\times\mathbb{C}^k) \cong X\times\mathbb{C}^{2N+k}$$
But why does there exists such $k$?



Equivalence of families indexes of Fredholm operators
Notice that $V^\perp\cap\ker T\_x^\*=0$ for every $x$, because for every $w\in V^\perp\cap\ker T\_x^\*$, $u\in H$, and $v\in V$, one has
$$\langle w,T\_x(u)+v \rangle = \langle w,T\_x(u) \rangle = 0.$$
By composing two isomorphisms $\ker P\_{V^\perp}T \ni u \mapsto u\oplus T(-u) \in \ker T^V$ and $\ker P\_{V^\perp}T\cong H/T^\*(V^\perp)$, one sees $\mathrm{Ind}\_2(T)=-\mathrm{Ind}\_1(T^\*)$.
Thus the quality of two indices follows from the fact that self-adjoint Fredholm operator
$\left[\begin{smallmatrix} & T^\*\\ T & \end{smallmatrix}\right]$ has index zero in either definition.


