# Strict transform does not modify the Normalization

I have a question about a reduction step in proof of Lemma 10.1.24 from Qing Liu's Algebraic Geometry and Arithmetic Curves on page 463:

Firstly we reduce to $$S$$ local, but then we replace $$\mathcal{C}$$ by it's desingulariation (which exists). Reason: because replacing $$\Gamma$$ by it's strict transform does not modify it's normalization.

Question/Problem: I don't completely understand the last reduction argument. $$\Gamma$$ was assumed as an irreducible component of closed fiber $$C_s$$ of structure map $$\mathcal{C} \to S$$ and it's normalization $$\Gamma'$$ is smooth.

Let denote by $$\pi:\mathcal{D} \to \mathcal{C}$$ the desingularization of $$\mathcal{C}$$, by $$T \subset \mathcal{D}$$ the strict transform of $$\Gamma$$ and by $$T'$$ normalization of $$T$$. Why $$T' \cong \Gamma'$$?

The composition $$T' \xrightarrow{n_T} T \hookrightarrow \mathcal{D} \xrightarrow{\pi} \mathcal{D}$$ factorize over $$\Gamma$$ thus we obtain a birational map $$b: T' \to \Gamma$$. Since $$T'$$ by construction is normal we use iniversal property of normalization map $$n_{\Gamma}: \Gamma' \to \Gamma$$ and obtain a birational map $$f: T' \to \Gamma'$$ of regular curves. Why is $$f$$ an isomorphism? That is why as Liu said the strict transform does not modify the normalization of $$\Gamma$$?