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I want to prove proposition 5.4 in Lawvere and Rosebrugh "Sets for Mathematics" but it seems I get a little stuck, when I try to mimic the approach for the covariant version in prop. 5.2.

My approach thus far goes as follows:

$\beta: B_2 \rightarrow B_1$ induces a $Y^\beta:Y^{B_1} \rightarrow Y^{B_2}$ since $Y^\beta = \ulcorner \text{eval}_1 \circ 1 \times \beta \urcorner$ is the wanted map, where $1= 1_{Y^{B_1}}$ and has $\text{eval}_2(Y^\beta \times 1_{B_2}) = \text{eval}_1 \circ 1 \times \beta$ by construction. To prove $Y^\beta \ulcorner g \urcorner = \ulcorner g\beta \urcorner$ for all $g:B_1 \rightarrow Y$ it is enough to prove that $\text{eval}_2 (Y^\beta \ulcorner g \urcorner \times 1_{B_2})= g\beta$ since by definition $\ulcorner g\beta \urcorner$ is unique with this property.

Now for calculations notice that:

$Y^\beta \ulcorner g \urcorner \times 1_{B_2}= (Y^\beta \times 1_{B_2})(\ulcorner g \urcorner \times 1_{B_2})$ by functoriality of products.

Calculating gives now:

$\text{eval}_2 (Y^\beta \ulcorner g \urcorner \times 1_{B_2}) = \text{eval}_2 ((Y^\beta \times 1_{B_2})(\ulcorner g \urcorner \times 1_{B_2})) = (\text{eval}_2(Y^\beta \times 1_{B_2}))(\ulcorner g \urcorner \times 1_{B_2}) = (\text{eval}_1 \circ 1 \times \beta)(\ulcorner g \urcorner \times 1_{B_2}) = \text{eval}_1 ((1 \times \beta)(\ulcorner g \urcorner \times 1_{B_2})) = \text{eval}_1 (\ulcorner g \urcorner \times \beta) $

And this is where I get stuck since $\text{eval}_1 (\ulcorner g \urcorner \times \beta)$ is not obviously equal to $ g \beta $, which would prove the equality in question. Okay continuing a bit further gives:

$\text{eval}_1 ((\ulcorner g \urcorner \times 1_{B_1})(1_1 \times \beta)) = (\text{eval}_1(\ulcorner g \urcorner \times 1_{B_1}))(1_1 \times \beta) = g (1_1 \times \beta)$

But then there is still the missing part of $1_1 \times \beta$ left, which I can't get rid of by simply using the definition of function spaces.

The operation $\ulcorner g \urcorner$ above denotes the unique $\ulcorner g \urcorner: 1 \rightarrow Y^{B_1}$ coming from any $g: 1 \times B_1 \rightarrow Y$, where $1$ is the terminal object in $\textit{Set}$.

EDIT: The definition of function space I use is the following.

Definition: $Y^B$ is a set together with an evaluation $\text{eval}:Y^B \times B \rightarrow Y$ such that for all $X$ and $f:X \times B \rightarrow Y$ there is a unique $\ulcorner f \urcorner: X \rightarrow Y^B$ such that $\text{eval}(\ulcorner f \urcorner \times 1_B)= f$.

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    Could you explain what $\ulcorner$ and $\urcorner$ mean? – 2017-02-23
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    It should be the unique induced operation on function sets coming from the exponentiation. – 2017-02-23
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    Firstly it should be $Y^\beta : Y^{B_1}\to Y^{B_2}$ and not $Y^\beta : Y^B_1\to Y^B_2$. Secondly, unless $1\times B_1 = B_1$ (which is an unnecessary assumption) $\text{eval}_1 (\ulcorner g \urcorner \times 1) = g\pi_2$ (where $\pi_2$ is the second projection of the product which is an isomorphism) and similarly we need to show that $\text{eval}_2 (Y^\beta \ulcorner g \urcorner \times 1_{B_2})= g\beta\pi_2$. Finally just note that $\text{eval}_1 (\ulcorner g \urcorner \times \beta) = \text{eval}_1 (\ulcorner g \urcorner \times 1)(1\times \beta) = g \pi_2 (1\times \beta) = g \beta \pi_2$. – 2017-02-25
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    @Nex how do you get that $\text{eval}_1(\ulcorner g \urcorner \times 1)=g\pi_2$? I don't recall that as part of the definition of the exponentiation? – 2017-02-25
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    Well, $\text{eval}_1$ is a morphism from $Y^{B_1}\times B_1$ to $Y$ and $\ulcorner g \urcorner \times 1$ is a morphism from $1\times B_1$ to $Y^{B_1} \times B_1$ so their composite $\text{eval}_1( \ulcorner g \urcorner \times 1)$ is a morphism from $1\times B_1$ to $Y$ and hence can't equal $g$ which is a morphism $B_1$ to $Y$, unless it is assumed that $1\times B_1 = B_1$ (in which case the morphisms $\pi_2$ will be identity morphisms). – 2017-02-25
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    Well, can't I just say that since $1 \times B_1 \cong B_1$ via an isomorphism this holds? – 2017-02-25
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    @Nex I think there is some misunderstanding about how the definition of exponentiation is used so I will just add this to the description. – 2017-02-25
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    $\pi_2$ is the isomorphism $1\times B_1 \to B_1$, and when you are calculating it seems to me necessary to take care of it. – 2017-02-25
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    Yes, it works for me now, thanks again! – 2017-02-25

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