Let $A$ denote an algebra finite dimensional, basic, and connected algebra over a algebraically closed field $K$. We denote by $mod A$ the abelian category whose objects are finitely generated right modules over $A$.


Let $T$ a tilting module. Denote the torsion pair induced by $T$ in $mod A$ by $(\mathcal{T}(T),\mathcal{F}(T))$ (<http://en.wikipedia.org/wiki/Tilting_theory>).


The book "Elements of the Representation Theory of Associative Algebras: Volume 1: Techniques of Representation Theory"
(<http://books.google.com.br/books/about/Elements_of_the_Representation_Theory_of.html?id=ayNHpi3tYhQC&redir_esc=y>) has an exercise that is hard to do:


If $N$ is an object of the $\mathcal{F}(T)$, then
$$
pd \; Ext^1\_A (T,N) \leq 1 + max(1, pd \; N).
$$
Anyone have an idea?


Ps.: $pd$ is the projective dimension.



Torsion pairs and projective dimension
In this answer I freely use results and notation from the book you refer to. Let me know if I need to fill in more details.


Let $U=\operatorname{Ext}^1\_A(T,N)$. We can assume $\operatorname{pd}\_B (U)=n>2$ since the statement holds trivially for smaller values. Let 
$$0 \rightarrow \Omega\_B U \rightarrow P\_B \rightarrow U \rightarrow 0$$
be the first step in a projective $B$-resolution of $U$. 


Since $N \in \mathcal F(T)$, we have $U \otimes\_B T=0$ and $\operatorname{Tor}^B\_1(U,T) \cong N$ as $A$-modules. Therefore, when we apply $- \otimes\_B T$ to the sequence above, we obtain an exact sequence of $A$-modules 
$$0 \rightarrow N \rightarrow \Omega\_B(U) \otimes\_B T \rightarrow T' \rightarrow 0,$$
where $T'$ is in $\operatorname{add}T$.


The module $\Omega\_B(U)$ is in the category $\mathcal Y(T)$ since it is a submodule of a projective $B$-module. There is an exact equivalence $- \otimes\_B T \colon \mathcal Y(T) \rightarrow \mathcal T(T)$, and therefore $$\operatorname{pd}\_A (\Omega\_B(U) \otimes\_B T) \geq \operatorname{pd}\_B (\Omega\_B(U))=n-1>1.$$ Since $\operatorname{pd}\_A (T') \leq 1$, it follows from the exact sequence that
$$\operatorname{pd}\_A(N)=\operatorname{pd}\_A (\Omega\_B(U) \otimes\_B T).$$
Combining with the previous inequality we get
$$\operatorname{pd}\_A(N) \geq \operatorname{pd}\_B (U)-1$$ which proves the statement.


