The dimension of kernels of a continuous family of Fredholm operators is upper semi continuous There is this question about continuous a family Fredholm Operators which I have been stuck at for a long time. It might be a straight forward result but I am unable to get to it. Suppose there is a continuous map,
$$f:X\rightarrow \mathcal{F}$$ 
where $\mathcal{F}$ deontes the space of all Fredholm Operators. I want to show that the function $\dim \ker f(x)$ is upper semi continuous, meaning that for every $x_0\in X$ there is a neighbourhood $U$ of $x_0$ such that 
$$\dim \ker f(x_0)\geq \dim \ker f(x)$$ 
for all $ \ x\in U$
I guess that a similar result would be true for the dimension of the cokernels as well, they will be lower semi continuous. Can someone help me with a proof of this.
The question was edited after @grew's comment.
 A: I think this is a corollary of the proof that the index is locally constant on the space of Fredholm operators.
Let $T: X \to Y$ be a Fredholm operator between Banach spaces. Since $T(X)$ is the image of a continuous linear operator and has finite codimension it will be closed. Hence we can choose a complement $W$ for $T(X)$ and we can also choose a complement $V$ for $\ker T$.
We now define for any continuous linear operator $S: X \to Y$ a linear operator $\widetilde{S}: V \times W \to Y$ by $\widetilde{S}(v, w) = S v + w$. Here we equip $V \times W$ with the norm $\lVert(v,w)\rVert_{\infty} = \max\{\lVert v\rVert, \lVert w\rVert\}$. It is the straightforward to check that the assignment $S \mapsto \widetilde{S}$ is continuous. Note that
\begin{equation}
 \widetilde{S}|_{V \times \mathbf{0}} = S|_V
 %\quad\text{and}\quad \widetilde{S}(\mathbf{0} \times W) = W
 \tag{1}\label{eq}
\end{equation}
Now by construction we have that $\widetilde{T}$ is bijective and hence invertible, which implies (using the continuity of $S \mapsto \widetilde{S}$) that for all $S$ in a neighborhood of $T$ the continuous linear map $\widetilde{S}$ is invertible. By (\ref{eq}) this implies that $S|_V$ is injective, so
\begin{equation}
 \dim \ker S ≤ \operatorname{codim}_X V = \dim \ker T
\end{equation}
Since $S$ was an arbitrary element of a certain open neighborhood of $T$ this completes the proof.
